Anna has 7,209 cans of soup that need to be boxed. If she puts 9 cans of soup in 1 box, how many boxes will she need.

Answers

Answer 1

Answer:

801 boxes

Step-by-step explanation:

[tex]\frac{7209}{9}[/tex] = 801

Answer 2
Set up a proportion first.
If 9 cans in 1 box, then x boxes will be need for 7209 cans.

9x= 7209
9x/9= 7209/9
x= 801

801 boxes will be needed for 7209 cans

Related Questions

Can someone help me on this question?

Answers

Answer:

  see below for the chart

  64 pages in 1 hour

Step-by-step explanation:

You want to fill in the chart showing the analysis of the problem statement saying HP can read 16 pages in 1/4 hour. The problem statement is requesting the number of pages he can read in 1 hour.

Chart

See below for a filled-in chart.

Solution

Analysis of the bottom line tells you that the right column is 4 times the middle column.

  1 = 4 × 1/4

  ?? = 4 × 16 = 64

Harry Potter can read 64 pages in one hour.

10. ANEW ACAR. Solve for x and CR. *
X=
Given: ANEW ACAR
NE = 11
EW = 12
I
AR = 4y - 12
CA = 4x + 3
NW = x + y

Answers

Applying the definition of congruent triangles, the values of x and y are:

x = 2; y = 6

Length of CR = 8 units.

How to Find  the Sides of Congruent Triangles?

If two triangles are congruent to each other, based on the CPCTC, all their corresponding angles, and their corresponding sides will be equal to each other.

Given that triangles NEW and CAR are congruent to each other, therefore:

NE = CA

NW = CR

EW = AR

Given the following measures:

NE = 11

EW = 12

AR = 4y - 12

CA = 4x + 3

NW = x + y

Therefore:

NE = CA

11 = 4x + 3

Solve for x:

11 - 3 = 4x

8 = 4x

8/4 = x

x = 2

EW = AR

12 = 4y - 12

12 + 12 = 4y

24 = 4y

24/4 = y

y = 6

NW = CR = x + y

Plug in the values of x and y:

CR = 2 + 6

CR = 8 units.

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Question 12 options:
Nick determines the remainder of

, using the remainder theorem.

How does she proceed to the correct answer?

Fill in the blanks from the work bank below: (do not enter spaces)

x = 1 x = -1 x = 0 -23 -19 -15 -29

Nick evaluates the numerator of the rational expression when

. He concludes that the remainder of the division is

Answers

The input value for the expression is 1 and the remainder is -15.

What is the remainder theorem?

The Remainder Theorem begins with an unnamed polynomial p(x), where "p(x)" simply means "some polynomial p with variable x". The Theorem then discusses dividing that polynomial by some linear factor x a, where an is simply a number.

Given polynomial is 5x⁴⁵ - 3x¹⁷ +2x⁴ - 19 and is divided by x + 1. Find the value of the polynomial at 1 and then calculate the remainder by synthetic solution.

F(x) = 5x⁴⁵ - 3x¹⁷ +2x⁴ - 19

F(1) = 5 - 3 + 2 - 19

F(1) = -19 + 4

F(1) = -15

The remainder will be calculated by synthetic calculation,

1 __5 ___-3___2_____-19

 |

 | ______5___ 2______4          

      5         2      4          -15

The remainder of the given expression is -15.

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The two-way frequency table contains data about students' preferred exercise.



Enjoys swimming Enjoys cycling Row totals

Likes running 28 62 90

Does not like running 46 64 110

Column totals 74 126 200


What is the joint relative frequency of students who do not like to run but enjoy cycling?

64%

55%

32%

23%

Answers

The joint relative frequency for those students who don't like to run, but however enjoy cycling, is C. 32%.

How to find the joint frequency ?

The joint frequency for students who like cycling, but do not like running, can be found by the formula :
= Number of students who enjoy cycling but don't enjoy running / Number of students in total

Number of students who enjoy cycling but don't enjoy running = 64

Number of students in total = 200

The joint frequency is :

= 64 / 200 x 100%

= 0.32 x 100 %

= 32 %

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Find the remainder term Rn in the nth order Taylor polynomial centered at a for the given function. Express the result for a general value of n. f(x)=e^-2x, a=2 Choose the correct answer below. O A. Rn (x) = (-2)^n e^-2c/(n+1)! (x- 2)^n for some c between x and 2. O B. Rn (x) = (-2)^n+1 e^-2c/(n+ 1)! for some c between x and 2. O C. (-2)1+1e 2c Rn(x)=?(n+1)!-(x-2)n + 1 for some c between x and 2. O D. n -2c OD. (x-2)"+1 for some c between x and 2.

Answers

The remainder term Rn is equal to (-2) raised to the power of n multiplied by e raised to the power of -2c, divided by n+1 factorial, multiplied by (x-2) raised to the power of n, for some c between x and 2.

The remainder term Rn in the nth order Taylor polynomial centered at a for the given function f(x)=e^-2x, a=2 is equal to (-2) raised to the power of n multiplied by e raised to the power of -2c, divided by n+1 factorial, multiplied by (x-2) raised to the power of n, for some c between x and 2. This remainder term represents the difference between the approximate value of the function given by the Taylor polynomial and the actual value of the function. By taking the limit as n approaches infinity, the remainder term can be used to determine the exact value of the function. The remainder term is an important concept in calculus as it is used to determine the accuracy of a given approximation.

For n=5, R5(x) = (-2)^5 e^-2c/(5+1)! (x-2)^5 for some c between x and 2.

Substituting n=5, x=7 and c=4, we have:

R5(7) = (-2)^5 e^-8/(6)! (7-2)^5

R5(7) = (-2)^5 e^-8/(720) (25)

R5(7) = (-2)^5 e^-8/18000

R5(7) = (-2)^5 e^-8/18000

R5(7) = -1.388888888 x 10^-9

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The square of a number is equal to 72 more than the number. Find all such numbers.

Answers

Answer:

You just need to think about this a little. There is a number, that you don't know, x, and if you square it ([tex]x^2[/tex]), that value will be equal to 72 more than the unknown number itself, so you put that into an equation.

Step-by-step explanation:

[tex]x^2=x+72\\x^2-x-72=0\\[/tex]

Apply quadratic formula (you can look it up)

[tex]x_1=\frac{-(-1)+\sqrt{(-1)^2+(-4*1*-72} )}{2(1)} \\\\x_2=\frac{-(-1)-\sqrt{(-1)^2+(-4*1*-72} )}{2(1)} \\\\[/tex]

There are two answers to the quadratic formula.

The rest should be easy.

Clarks Inc., a shoe retailer, sells boots in different styles. In early November the company starts selling “SunBoots” to customers for $70 per pair. When a customer purchases a pair of SunBoots, Clarks also gives the customer a 30% discount coupon for any additional future purchases made in the next 30 days. Customers can’t obtain the discount coupon otherwise. Clarks anticipates that approximately 20% of customers will utilize the coupon, and that on average those customers will purchase additional goods that normally sell for $100. Required: 1. How many performance obligations are in a contract to buy a pair of SunBoots? 2. Assume Clarks cannot estimate the standalone selling price of a pair of SunBoots sold without a coupon. Prepare a journal entry to record revenue for the sale of 1,000 pairs of SunBoots.

Answers

The journal entry is given in answer part b

What is Discount coupon?

A discount coupon is defined as a certificate that entitles the customer to avail of a reduction in the invoice price if the terms mentioned on it are met. It is a technique applied by the sellers to expand sales.

Given that, Clarks Inc., a shoe retailer, sells boots in different styles. In early November, the company starts selling “SunBoots” to customers for $70 per pair. When a customer purchases a pair of SunBoots, Clarks also gives the customer a 30% discount coupon for any additional future purchases made in the next 30 days. Customers can’t obtain the discount coupon otherwise. Clarks anticipates that approximately 20% of customers will utilize the coupon, and that on average those customers will purchase additional goods that normally sell for $100.

a) Two performance obligations are associated with contract. The first one is related to the sun boots, and the other one is associated with discount coupon provided. The discount coupon would be considered a separate obligation because it gives a certain right to the customers that they would not have received if the sale was not made to them.

b) Journal entry =

Account title and explanation            Debit ($)                    Credit ($)

Cash (1000×$70)                                    $70000                  

Sales Revenue ($70,000-$6000)                                           $64000

Deferred Revenue                                                                    $6000

Now,

Computation of the deferred revenue:

Deferred revenue = pairs sold × Average purchase price × Discount × Redemption of coupon

= 1000 × 100 × 30% × 20%

= $6000

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Which of the following represents members of the domain of the graphed function?

See attached picture

Responses

{-4, 2, 3}
{-4, 0, 1}
{1, 2, 3, 4}
{1, 2, 3, 5}

Answers

The {-4, 0, 1} represents members of the domain of the graphed function.

What is the domain of the function?

A function is a mathematical object that accepts input, appears to apply a rule to it, and returns the result.

A function can be thought of as a machine that requires in a number, performs some operation(s), and then outputs the result.

The domain of a function is the collection of all its inputs. Its codomain is the set of possible outputs.

The range refers to the outputs which are actually used.

Domain: {-4, 0, 1}.

Simply list the domain as -4 < x < 2, which would imply ALL values between -4 and 2 inclusive.    

Yes, this is a function. No x-values repeat, and it passes the Diagonal Line Test for functions.

Hence, the {-4, 0, 1} represents members of the domain of the graphed function.

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Here we show that function defined on an interval value property cannot have (a; b) and satisfying the intermediate removable or (a) jump discontinuity. Suppose has & jump discontinuity at Xo € (a,b) and lim f (x) lim f (x) xx0 {ix0 Choose 0 such that lim f (x) < 0 < lim f (x) and 0 + f(xo) xI*o Xx0 In Exercise & we showed there is interval [xo 0,.Xo) such that f(x) < 0 if Xe [xo 6,xo): Likewise, there an interval (xo, Xo + 6] such that f(x) > 0 if xe(xo, Xo + 6]. Conclude that does not satisly the intermediate value property on [xo 6,xXo + 6]. (6) Suppose has a removable discontinuity at Xo € (a,b) and a = lim f(x) < f(xo) Show that there is an interval [xo = 6,Xo) such that f(x)< a+[f(xo) - &] if x e[xo 6,Xo]: Conclude that f does not satisfy the intermediate value property

Answers

f cannot have a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex] and  [tex]$$ \lim _{x \uparrow x_0} f(x) < \lim _{x \mid x_0} f(x) .$$[/tex]

f cannot have a removable discontinuity at [tex]$$x_0 \in(a, b) $$[/tex] and [tex]\alpha=\lim _{x \rightarrow x_0} f(x) < f\left(x_0\right)[/tex]

Let f be a function defined on (a, b) satisfies intermediate value property.

Claim: f ca not have removable on jump discontinuity.

Suppose f has a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex]

We take [tex]$\theta$[/tex] such that

[tex]$$\lim _{x \rightarrow x_0} f(x) < \theta < \lim _{x \downarrow x_0} f(x) \text { and } \theta \neq f\left(x_0\right)$$[/tex]

Now there exist [tex]$\delta > 0$[/tex] such that [tex]$f(x) < \theta$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right)$[/tex] and [tex]$f(x) > \theta$[/tex] for all [tex]$x \in\left(x_0, x_0+\delta\right]$[/tex]

Now [tex]$f\left(x_0-\delta\right)[/tex][tex]< \theta < f\left(x_0+\delta\right)$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0+\delta\right] \backslash\left\{x_2\right\}$[/tex] and [tex]$f\left(x_0\right) \neq \theta$[/tex].

Therefore the point [tex]$\theta$[/tex] has no preimage under f

that is, there does not exists [tex]$y \in\left[x_0-\delta, x_0+\delta\right][/tex] for which

[tex]$$f(y)=\theta[/tex] because [tex]\left\{\begin{array}{l}y=x_0 \Rightarrow f(y) \neq \theta \\y > x_0 \Rightarrow f(y) > \theta \\y < x_0 \Rightarrow f(y) < \theta\end{array}\right.$$[/tex]

Therefore f does not satisfies intermediate value property on [tex]$\left[x_0-\delta, x_0+\delta\right]$[/tex],

Hence f does not satisfies IVP on (a, b) which is not possible because we assume f satisfies IVP on (a, b),

Therefore f can not have a jump discontinuity.

Suppose f has a removable point of discontinuity at [tex]$x_0 \in(a, b)$[/tex],

Let [tex]$\alpha=\lim _{\alpha \rightarrow x_0} f(x)$[/tex],

Let [tex]\alpha < f\left(x_0\right)$[/tex] so [tex]$f\left(x_0\right)-\alpha > 0$[/tex].

Now [tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex] then [tex]\exists$ \delta > 0$[/tex] such that

[tex]$$\begin{aligned}& |f(x)-\alpha| < \frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left\{x_0-\delta, x_0-\alpha\right]-\left\{x_0\right\} \\& \Rightarrow \quad f(x) < \alpha+\frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left[x_0-\delta, x_0+\delta\right]-\left\{x_0\right\}\end{aligned}$$[/tex]

So [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2}$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right]-\left\{x_0\right\}$[/tex]

Now [tex]$f\left(x_0\right) > \alpha$[/tex].

And  [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2} < f\left(x_0\right)$[/tex] for all [tex]$x \in\left[\left(x_0 \delta, x_0\right)\right.$[/tex]

Let [tex]$\mu=\frac{f\left(x_0\right)+\alpha}{2}$[/tex].

Then there does not exist [tex]$e \in\left[x_0-\delta, c\right]$[/tex] such that [tex]$f(c)=\mu$[/tex]

Because for [tex]$e=x_0 \quad f(e) > \mu$[/tex]

                for [tex]$c < x_0 \quad f(c) < \mu$[/tex].

Therefore f does not satisfy IVP on [tex]$\left[x_0-\delta_1 x_0\right]$[/tex] which contradict our hypothesis,

therefore [tex]$\alpha \geqslant f\left(x_0\right)$[/tex]

Let [tex]$\alpha > f\left(x_0\right)$[/tex]. so [tex]$\alpha-f\left(x_0\right) > 0$[/tex]

[tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex]

Then [tex]\exists $ \varepsilon > 0$[/tex] such that

[tex]$|f(x)-\alpha| < \frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$\left.x \in\left[x_0-\varepsilon_0 x_0+\varepsilon\right]\right\}\left\{x_i\right\}$[/tex]

[tex]$\Rightarrow f(x) > \alpha-\frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0+\varepsilon\right] \backslash\left\{x_0\right\}$[/tex]

[tex]$\Rightarrow f(x) > \frac{\alpha+f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0\right)$[/tex]

Now [tex]$f\left(x_0\right) < \alpha$[/tex]

The [tex]$f(x) > \frac{f\left(x_0\right)+\alpha}{2} > f\left(x_0\right)$[/tex].

So [tex]$f\left(x_0\right) < \frac{f\left(x_0\right)+\alpha}{2} < f(x)$[/tex] for all [tex]$x \in\left[x_0 \varepsilon, \varepsilon_0\right)$[/tex]

Let [tex]$\eta=\frac{f\left(x_e\right)+\alpha}{2}$[/tex]

Then there does not exist [tex]$d \in\left[x_0-\varepsilon, x_0\right]$[/tex] such that [tex]$f(d)=\xi$[/tex].

Because if [tex]$d=x_0, f(d)=f\left(x_0\right) < \eta$[/tex] if [tex]$d E\left[x_0-\varepsilon, x_0\right)$[/tex]

Then [tex]$f(d) > \eta$[/tex]

Therefore f does not satisfies IVP on [tex]$\left[x_0-\varepsilon, x_0\right]$[/tex] which contradict olio hypothesis.

Therefore [tex]$\alpha \leq f\left(x_0\right)$[/tex] (b) From (a) and (b) it follows [tex]$\alpha=f\left(x_0\right)=\lim _{x \rightarrow x_0} f(x)$[/tex]. Therefore f can not have a removable discontinuous

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If (x+2) is a factor of

Answers

Answer: a=3 and b=1

Step-by-step explanation:

Let p(x)=x 2 +ax+2b

If x+2 is a factor of p(x)

Then by factor theorem

p(−2)=0

⇒(−2)

2

+a(−2)+2b=0

⇒4−2a+2b=0

⇒a−b=2   ---(i)

Also given that a+b=4    ---(ii)

Adiing (i) and (ii) we get

2a=6

⇒a=3

putting a=3 in (i) we get

a−b=2

⇒3−b=2

⇒b=1

Therefore we get a=3 and b=1

A circle with center w is shown in the figure below.

Answers

Radius = TW

Diameter = TX

Chord = UV

The length of TX is 6 units.

What is a circle?

A circle is a two-dimensional figure with a radius and circumference of 2 x pi x r.

The area of a circle is given as πr².

We have,

From the figure:

Diameter = TX

Radius = TW or WX or WY

Chord = UV

WY = 3 units

TX = TW + WX = 3 + 3 = 6 units.

Thus,

TX is 6 units.

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a duck flew at 18 miles per hour for 3 hours than at 15 miles per hour for 2 ours how far did the duck fly in all

Answers

The distance the duck fly in all is 99 miles

How to determine how far the duck fly in all

From the question, we have the following parameters that can be used in our computation:

Distance 1: 18 miles per hour for 3 hoursDistance 2: 15 miles per hour for 2 hours

The distance covered in all can be calculated as

Distance = The sum of the product of speed and time

Substitute the known values in the above equation, so, we have the following representation

Distance = 18 * 3 + 15 * 2

Evaluate the products

This gives

Distance = 54 + 45

Evaluate the sum

Distance = 99 miles

Hence, the distance is 99 miles

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5x^2-13+6=0
i need helppp

Answers

Answer:

answer is 2.5

Step-by-step explanation:

Write an equation for the line through the given point with the given slope in slope-intercept form. (10, –9); m = –2

Please explain.

Answers

Considering the definition of a line, the equation of the line that passes through the point (10, -9) and has a slope of -2 is y= -2x +11.

Linear equation

A linear equation o line can be expressed in the form y = mx + b

where

x and y are coordinates of a point.m is the slope.b is the ordinate to the origin and represents the coordinate of the point where the line crosses the y axis.Line in this case

In this case, you know:

The line has a slope of -2.The line passes through the point (10, -9).

Substituting the value of the slope m and the value of the point y=mx+b, the value of the ordinate to the origin b is obtained as follow:

-9= -2×10 + b

-9= -20 + b

-9 + 20= b

11= b

Finally, the equation of the line is y= -2x +11.

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You buy 9 granola bars for a camping trip. Each bar costs $0.95. What
is the total cost of 9 granola bars before tax?

Answers

Answer:

Step-by-step explanation: 9 multiplied by .95 = $8.55 before tax

what does
QR=
PR=
M

Answers

We need to use Right Angled Triangle Properties.

PR≈15 and QR≈12 is the length of the sides.

What is a Right Angled Triangle?

According to the definition of a right triangle, a triangle is referred to as a right-angled triangle or simply a right triangle if one of its angles is a right angle (90o). Triangle PQR is a right triangle in the illustration, and its components are its base, height, and hypotenuse. Here, QR is the base, PQ is the height, and PR is the hypotenuse. The essential side of a right triangle is its hypotenuse, which is also its largest side and sits across from the triangle's right angle.

In a right triangle we have: (Hypotenuse)2 = (Base)2 + (Altitude)2

Area of a right triangle = (1/2 × base × height) square units.

Calculation:

PR is the Hypotenuse of the Triangle.

PQ=9

Angle ∡PQR=37° and Angle ∡RPQ=53°

So Sin(theta)=opposite side/Hypotenuse

⇒Sin(37°)=9/Hypotenuse

⇒Hypotenuse=9/0.6018

⇒Hypotenuse=14.955

PR⇒14.955

We know that (Hypotenuse)2 = (Base)2 + (Altitude)2

⇒(14.955)2=(9)2+(QR)2

QR=11.9438

PR≈15 and QR≈12

We need to use Right Angled Triangle Properties.

PR≈15 and QR≈12 is the length of the sides.

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Sample Response: First, I wrote each ratio as a fraction. Then I found a common denominator of 20. The fraction 4/10 is 8/20. Comparing numerators, 8 is less than 12, so the ratio 4:10 is smaller.

Which did you include in your answer?
Check all that apply.
Write the ratios as fractions.
Find a common denominator.
Compare the numerators to see that 4:10 is the smaller ratio.

Answers

In the aforementioned scenario, the choice you must include in your response is to find a common denominator.

In the simplest terms, what is fraction?

One can refer to a fraction as being in its lowest term or in its simple form when the numerator and denominator are known to have no common factors besides 1. In general, any number of equal pieces is represented by a fraction, which is a portion of the entire. A fraction, such as one-half, eight-fifths, or three-quarters, indicates the number of parts of a particular size when it is used in daily speech.

The common denominator will be between 10 and 20, and there is a common numerator as well. The common denominator will be 20 because both can divide it and 10 is less than 20.

Finding a common denominator is the solution you should choose in the scenario mentioned above.

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Staci pays $32.70 for 5 cell phone cases. Each case costs the same amount. How much does each case cost?

Part A
Which expression represents the problem?
$32.70 × 5
$32.70 ÷ 5
$32.70 + 5
$32.70 – 5
Part B
Evaluate the expression from Part A.

$ ( ??? )

Answers

Answer:

$32.70 ÷ 5

Each case is $6.54.

Step-by-step explanation:

$3270 for 5 cases, meaning you would split $32.70 into 5.

$32.70 ÷ 5 phone cases = $6.54

This also means each case is $6.54. To prove this, multiply by 5.

Find f. (Use C for the constant of the first antiderivative and D for the constant of the second antiderivative.)
f ''(x) = 2x + 4ex

Answers

After integration, the required function f is (2x³ - sin (x) + Cx + D).

What is the integration of 'xⁿ' and 'sin (x)'?

[tex]\int {x^{n} } \, dx = \frac{x^{n+1} }{n+1} + C\\\\\\\int {sinx} \, dx = -cosx + C[/tex]

Given, f''(x) = 12x + sin x

Therefore,

[tex]\int {f''(x)} \, dx \\\\=\int {f'(x)} \, dx \\\\\\= \int{12x + sin x} \, dx + C\\\\= 6x^{2} - cosx + C\\[/tex]

Again, f'(x) = 6x - cos (x) + C

Therefore,

[tex]\int {f'(x)} \, dx\\ \\=\int {f(x)} \, dx \\\\= \int {6x^{2} - cosx + C } \, dx \\\\= 2x^{3} - sinx + Cx + D[/tex]

Therefore, the required function is (2x³ - sin (x) + Cx + D).

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Find the Derivative of 1/x+2

Answers

Answer: 2x+1

                  x

Step-by-step explanation:

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15
minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.
Little's law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84
shoppers per hour make a purchase and each of these shoppers spends an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers,
on average, are waiting in the checkout line to make a purchase at the Good Deals Store

Answers

Note that at any time during business hours, about 7 shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store. This is solved using Little Law.

What is littles Law?

Little's Law asserts that the long-term average number of individuals in a stable system, L, is equal to the long-term average effective arrival rate,λ, multiplied by the average duration a customer spends in the system, W.

To compute,

First, let's make sure that all the variables use the same time unit.

Thus, if there are 84 shoppers who are making a purchase per house, then there will be: 84/60 minutes

= 1.4

This means that  λ (rate) = 1.4 and the average time (W) is 5 minutes

Using Little's Law which states that the queuing formula is:

L = λW

Where L = Average number of factors in the system (in this case, people on the queue)λ = Average arrival and departure rate; andW = lead time that a factor spends in the system

Note that:

λ = 1.4 (computed)

W = 5 mintues (given)

Hence, the Average number of people in queue per time is:

L = 1.4 x 5

L = 7 shoppers.

Thus, it is correct to state that at any time during work hours, about 7 shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store.

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Full Question:

If shoppers enter a store at an average rate of r shoppers per minute and each stays in the store for an average time of T minutes, the average number of shoppers in the store, N, at any one time is given by the formula N=rT. This relationship is known as Little's law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.

Little's law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spend an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

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The expoential function given by f(x) = e^x called the (a. natural logarithmic, natural exponential, c. 1 to 1 exponential, d. 1 to 1 algebraic, e. transcendental algebraic) function and the base e is called the (a. algebraic, b. 1 to 1, c. natural, d. rational, e. transcendental)

Answers

Answer:

natural exponential

natural

quizlet

Answer:

b. natural exponential

c. natural

Step-by-step explanation:

Given function:

[tex]f(x)=e^x[/tex]

The given exponential function is called the:

natural exponential function

The base e is called the:

natural base

The number "e" occurs naturally in math and the physical sciences.

It is the base rate of growth shared by all continually growing processes, and so is called the natural base.

It is an irrational number and named after the 18th century Swiss mathematician, Leonhard Euler, and so is often referred to as "Euler's number".

I need help
How do you find x:
2^(2x) + 2^(x+2) - 12 = 0

Answers

Answer:

[tex]x=1[/tex]

Step-by-step explanation:

[tex]2^{2x}-2^2 2^x-12=0 \\ \\ (2^x)^2-4(2^x)-12=0[/tex]

Factoring the quadratic in [tex]2^x[/tex],

[tex](2^x+6)(2^x-2)=0 \\ \\ 2^x=-6,2 \\ \\ 2^x=2 \implies x=1 \\ \\ 2^x=-6 \implies x \notin \mathbb{R} \\ \\ \therefore x=\log_{2} (6)[/tex]

i literally don’t know

Answers

∠U is congruent ∠S because all right angles are congruent.

Define congruent.

If it is possible to superimpose one geometric figure on the other so that their entire surface coincides, that geometric figure is said to be congruent, or to be in the relation of congruence. When two sides and their included angle in one triangle are equal to two sides and their included angle in another, two triangles are said to be congruent. This concept of congruence appears to be based on the idea of a "rigid body," which may be moved without affecting the internal relationships between its components.

Given

∠U ≅ ∠S

All right angles are congruent.

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Alex biked a 48-mile trail in 4 hours. For the first half of the time, he biked twice as fast as he did the second half of the time. What speed did Alex bike the first half?

Answers

Alex's speed in the first half of the time is 12 miles per hour

How t find the speed of biking

Information from the problem include:

For the first half of the time, Alex biked a 48-mile trail in 4 hours.

he biked twice as fast as he did the second half of the time.

Average speed is calculated using the ratio of distance to time spent

this is represented by the formula

= distance / time

The speed for the first half is  solved as below

= 48 mile / 4 hours

= 12 mile / hours

= 12 mph

speed for the first half is 12 mph

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Does anyone know the answer for this?

Answers

The Venn diagram has been drawn and the product of LCM and GCF are the same as 24 x 36.

What is a Venn diagram?

A Venn diagram is a pictorial representation of data represented in a circle.

The intersection parts of the circle represent the commonality of both sets.

As per the given, 24 and 36

The factors of 24 are as 2³ × 3

The factors of 36 are as 2² × 3²

The common factors 2² × 3

The LCM of 24 and 36 is 72 and the GCF is 12.

LCM x GCF = 24 x 12

72 x 12 = 24 x 12

864 = 864

Hence "After drawing a Venn diagram, it was discovered that the LCM and GCF product equals 24 x 36.".

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Convert the measurement. 28.4 grams ≈ ounces

Answers

Answer: 1.0017805194 ounces

Step-by-step explanation:

To change 28.4 g to oz divide the mass in grams by 28.349523125. The 28.4 grams to ounces formula is [oz] = [28.4] / 28.349523125. Thus, we get (all results rounded to 2 decimals):

28.4 g to oz = about 1 oz

28.4 g to ounces =  about 1 ounces

28.4 grams to ounces = about  1 oz

Twenty-eight point four grams to ounces equal 1 international avoirdupois ounces, the unit used to measure grocery products (mass) in retail stores in the United States for example.

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If x varies directly as y, find x when y = 8 a) x = 6 when y = 32 b) x = 14 when y = -2

Answers

a) The value of x is 42.6 when k =16/3 is directly proportional to y.

b) The value of x is -56 when k is -7  is directly proportional to y.

What kind of variation is one where x and y are directly proportional?

If x = ky for some constant k can be used to indicate the relationship between the variables y and x, then we may say that y varies directly with y or that x is directly proportional to y.

x varies directly as y

x α y

x = ky

a) x=6 ; y = 32

x = ky

6 = k(32)

k = 16/3

if y = 8

then,

x = (16/3) * 8

= 128/3

x = 42.6

Hence, the value of x is 42.6 when k =16/3 is directly proportional to y.

b) x= 14 and y = -2

x = ky

14 = k(-2)

k = -7

if y = 8

then,

x = ky

x= (-7) 8

x = -56

Hence, the value of x is -56 when k is -7  is directly proportional to y.

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which statement describes the solutions of this equation? x/x+2 + 1/x = 1

Answers

Answer:

[tex]\frac{x}{x+2}+\frac1x=1\\\rightarrow \frac{x^2+(x+2)}{(x^2+2x}=1\\\rightarrowx^2+x+2=x^2+2x\\\rightarrowx+2=2x\\\rightarrow x=\boxed2[/tex]

Only one solution, which is two. The answer should be C. You could obviously see that if there is no squaring then there will be no extraneous solutions.

Find the area of a circle with radius 6 ft.

Answers

Answer:

The area of the circle is 113.04 square feet.

Step-by-step explanation:

Given;Radius (r) = 6 ftπ = 3.14Formula;A = πr²

A = 3.14 × (6)² ft

A = 3.14 × 36 ft

A = 113.04 square feet.

Thus, The area of the circle is 113.04 square feet.

Answer:

28.274 square feet

Step-by-step explanation:

Multiply the radius by pi

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