What is the length of DE ?

Answer:

343

Step-by-step explanation:

How many three-letter ”words” can be made from 7 letters FGHIJKL” if repetition of letters (a) is allowed? (b) is not allowed? Solution: (a) If repetition is allowed each letter can be any of the 7. So number ways is 7 × 7 × 7=73 = 343

Answer: See below

Step-by-step explanation:

Given:

The letters FGHIJKL

To find:

The number of three letter words can be formed if repetition is allowed (or) if repetition is not allowed

[tex]$$(a) If repetition is allowed:Total no of three letter words can be made$$=7_{C 1} x 7_{C 1} x 7_{C 1}=7^{3}=343$$(b) If repetitions is not allowed :Total no of three letter words can be made without repetition is $$\begin{gathered}=7_{C 1} \times 6_{C 1} \times 5_{C 1} \\=7 x 6 \times 5=210\end{gathered}$$[/tex]

Answer:

Number 4

Standard form - Not sure if right

y = − 0.2734375x^2 − 1.09375x +11.40625

Factored form - Not sure if right

y= 0.0390625 (−7x^2 −28x+292)

Number 5

Either form of the equation other than standard form

Vertex form : y = −(x +3)^2 + 25

Vertex of the parabola

(-3,25)

X and Y intercepts

x-intercept(s): (2,0) ,(−8,0)

y-intercept(s): (0,16)

Answer:

that's all I could do. hope it helps.

good day.

Answer:

Step-by-step explanation:

1250x0.05x1.5=93.75

93.75 is the interest

1343.75 is the total ammount.

Answer:

Proof below

Step-by-step explanation:

If [tex]p[/tex] is prime such that [tex]p \neq2[/tex] and [tex]p \neq 3[/tex], then [tex]p[/tex] divided by 6 has remainder of 1 or 5.

Case 1: p/6 has remainder 0, 2, or 4.

Then there exists some natural number [tex]n[/tex] such that either [tex]p=6n[/tex], [tex]p=6n+2[/tex], or [tex]p=6n+4[/tex].  However,

By definition of divisibility, [tex]p[/tex] is divisible by 2.  Since [tex]p \neq2[/tex], [tex]p[/tex] cannot be prime.  This contradiction implies [tex]p[/tex] cannot have a remainder of 0, 2, or 4 when divided by 6.

Case 2: p/6 has remainder 3.

Then there exists some natural number [tex]n[/tex] such that [tex]p=6n+3[/tex].  However, if [tex]p=6n+3[/tex], then [tex]p=3*(2n+1)[/tex], so [tex]p[/tex] is divisible by 3.  Since [tex]p \neq 3[/tex], [tex]p[/tex] is not prime.  This contradiction implies [tex]p[/tex] cannot have a remainder of 3 when divided by 6.

Therefore, if [tex]p[/tex] is a prime number such that  [tex]p \neq2[/tex] and [tex]p \neq 3[/tex], then when divided by 6, [tex]p[/tex] must have a remainder of 1 or 5.

Check p=2

Consider [tex]p=2[/tex].  Then [tex]q=2+70=72[/tex].  However, [tex]72=2*36[/tex], so [tex]q[/tex] is not prime.  Hence, [tex]p \neq2[/tex].

Check p=3

Consider [tex]p=3[/tex].  Then [tex]q=3+70=73[/tex], and [tex]r=73+70=143[/tex].  However, [tex]143=11*13[/tex], so [tex]r[/tex] is not prime.  Hence, [tex]p \neq 3[/tex].

So, thus far, we've proven that if there is a prime trio with a common difference of 70, that [tex]p \neq2[/tex] and [tex]p \neq 3[/tex].

Proof of the rest of primes by contradiction

By way of contradiction, assume that there does exist some prime trio [tex]\{ p,q,r \}[/tex] such that [tex]q-p=r-q=70[/tex], and [tex]p \neq2[/tex] and [tex]p \neq 3[/tex].

Then, by the Lemma proven earlier, when [tex]p[/tex] is divided by 6, it must have a remainder of 1 or 5.

Case 1: p/6 has remainder 5

Assume that [tex]p[/tex] has remainder 5 when divided by 6.  Then, there exists some natural number [tex]n[/tex], such that [tex]p=6n+5[/tex].  

[tex]q=p+70\\q=(6n+5)+70\\q=6n+75\\q=3*(2n+25)[/tex]

Since [tex]n[/tex] was a natural number, and the natural numbers are closed over multiplication and addition (meaning, multiplying and adding more natural numbers results in another natural number), then [tex]q[/tex] is divisible by 3.

Since [tex]p[/tex] is prime, [tex]0 < p[/tex], which implies [tex]0+70 < p+70[/tex]

Since [tex]q=p+70[/tex], [tex]70 < q[/tex], so [tex]q[/tex] cannot equal 3.  Since [tex]q[/tex] is divisible by 3, but [tex]q \neq 3[/tex], [tex]q[/tex] is not prime, which is a contradiction to the existence of this prime trio.

Therefore, either [tex]p[/tex] cannot have a remainder of 5 when divided by 6, or the prime trio does not exist.

Case 2: p/6 has remainder 1

Assume that [tex]p[/tex] has remainder 1 when divided by 6.  Then, there exists some natural number [tex]n[/tex], such that [tex]p=6n+1[/tex].  

[tex]q=p+70\\q=(6n+1)+70\\r=q+70\\r=((6n+1)+70)+70\\r=6n+141\\r=3*(2n+47)[/tex]

Since [tex]n[/tex] was a natural number, and the natural numbers are closed over multiplication and addition (meaning, multiplying and adding more natural numbers results in another natural number), [tex]r[/tex] is divisible by 3.

Since [tex]q[/tex] is prime, [tex]0 < q[/tex], implying [tex]0 +70 < q+70[/tex].

Since [tex]r=q+70[/tex], then [tex]70 < r[/tex].  Therefore, [tex]r\neq 3[/tex].

Since [tex]r[/tex] is divisible by 3, but [tex]r\neq 3[/tex], [tex]r[/tex] is not prime, which is a contradiction to the existence of this prime trio (with a common difference of 70).  Therefore, either [tex]p[/tex] cannot have a remainder of 1 when divided by 6, or the prime trio does not exist.

Since [tex]p[/tex] was prime, by our lemma [tex]p[/tex] must have either a remainder of 1 or 5.  Since both remainder possibilities resulted in a contradiction, our contradiction assumption is false, so there cannot exist a prime trio such that their common difference is 70.

A mean is an arithmetic average of a set of observations. The score that Taylor needs to make in her eleventh test for the average to be 87 is 93.

A mean is an arithmetic average of a set of observations. it is given by the formula,

Mean = (Sum of observations)/Number of observations

The score earned by Taylor in her first 10 test are 78, 92, 98, 87, 86, 72, 92, 81, 86, 92. Also, she wants the average to be 87, therefore, we can write,

(78 + 92 + 98 + 87 + 86 + 72 + 92 + 81 + 86 + 92 + x)/11 = 87

864 + x = 957

x = 93

Hence, The score that Taylor needs to make in her eleventh test for the average to be 87 is 93.

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The equation in slope intercept form is y = - 1 / 4 x + 1

Using slope intercept form,

y = mx + b

where

Therefore,

m = - 1 / 4

it passes through (8, -1)

Hen e,

-1 = - 1 / 4 (8) + b

-1  = - 2 + b

b = -1 + 2

b = 1

Therefore,

y = - 1 / 4 x + 1

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Answer:

-2 ≤ x ≤10

Step-by-step explanation:

Notice that x can never be less than -2  and can never exceed.

-2 and 10 are all inclusive  thus the last answer is not correct as it tries to leave them out

Answer:

p = 7.3

Step-by-step explanation:

Equation: p - 2.5 = 4.8

add 2.5 to both sides:

P - 2.5 + 2.5 = 4.8 + 2.5

simplify

P = 7.3

hope that helps! :)

The approximate 95 % confidence interval for p is found by the formula

p^ ±  Z∝/2 √p^q^/n

The sample size n can be found by the

The formula

n= (Z∝/2)² p^q^/e²

The approximate 95 % confidence interval for p is 0.165; 0.243

This indicates 95 % confidence interval level.

A sample of 6239 nonprofits should be surveyed

Solution

Part a:

Sample Proportion p^= 84/412= 0.2038 ≅ 0.204

q^ = 1-p^= 1.0 -0.204= 0.796

The degree of confidence is 95% so  z∝/2= 1.96

Putting the values

p^ ±  Z∝/2 √p^q^/n

0.204 ±  1.96 √0.204(0.796)/ 412

0.204 ± 0.0389

0.1650 ; 0.2429

0.165; 0.243

The approximate 95 % confidence interval for p is 0.165; 0.243

Part B

This indicates 95 % confidence interval level.

Part C

With ± 0.01 of 0.95  Confidence interval gives range  0.94 to 0.96

For 0.96  interval Z∝/2 = 2.054

For 0.94  interval Z∝/2 = 1.8808

and error  e= 0.01

For solving n  the formula is

n= (Z∝/2)² p^q^/e²

and e= 0.01

Putting the values

n= 1.96²(0.204)(0.796)/ 0.01²

n= 6238.1437≅ 6238.14

A sample of 6239 nonprofits should be surveyed.

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The amount of money a farmer can make by selling the milk produced by the cow in the month of April is $90

From the equation given, we have that

$3 = I gallon

Let's take it that the farmer sells one gallon by per day

We know that the month of April has 30 days

So if he makes $3 in 1 day

He would make x in 30 days

Collect like terms

$30 × 30 days = amount

Multiply through

Amount = $90

Thus, the amount of money a farmer can make by selling the milk produced by the cow in the month of April is $90

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Answer:

C

Step-by-step explanation:

(5r + 8 - 6r)(4 + 2r - 7) ← collect like terms inside each parenthesis

= (8 - r)(2r - 3)

each term in the second factor is multiplied by each term in the first factor.

8(2r - 3) - r(2r - 3) ← distribute parenthesis

= 16r - 24 - 2r² + 3r ← collect like terms

= - 2r² + 19r - 24

Answer:

11/12

Step-by-step explanation:

2 3/4 ÷ 3

11/4 ÷ 3

11/4 × 1/3

11/12

Answer:

Hello! The answer is 11/12.

Step-by-step explanation:

Convert improper fractions:

[tex]2\frac{3}{4} \\= \frac{11}{4}[/tex]

Change 3 to its reciprocal:

[tex]\frac{1}{3}[/tex]

Multiply:

[tex]\frac{11}{4} * \frac{1}{3} \\= 11/12[/tex]

Answer:

= -x+6

Step-by-step explanation:

we can simply substitute (x-4) in x place because when we say function of x , we are basically saying when we substitute a given value in the place of x we will find a given y.so for example if we have a question f(4), we can simply plug 4 to -x +2 so we will find y or f(x) = -2.

so the answer for the above question will be f(x)= -x +6.

Answer:

[tex]\huge\boxed{\sf \$ \ 1.47}[/tex]

Step-by-step explanation:

6 oranges = $0.49

Multiply 3 to both sides

6×3 oranges = $0.49 × 3

18 oranges = $1.47

[tex]\rule[226]{225}{2}[/tex]

Answer:

  about 33.3%

Step-by-step explanation:

The fraction of space not occupied will be the same as the unoccupied fraction of a cylinder of the same height and diameter as a sphere.

The volume of a cylinder of diameter d and height d will be ...

  V = πr²h = π(d/2)²(d) = (π/4)d³

The volume of a sphere of diameter d is ...

  V = 4/3πr³ = 4/3π(d/2)³ = (π/6)d³

The fraction of space that is occupied is ...

  occupied space = (sphere volume)/(cylinder volume)

  occupied space = ((π/6)d³) / ((π/4)d³) = 4/6 = 2/3

The fraction of unoccupied space is ...

  unoccupied fraction = 1 -2/3 = 1/3

  unoccupied fraction ≈ 33.3%

Answer:

Putangina okay

Step-by-step explanation:

#CarryOnLearning

Answer: B and D

Step-by-step explanation:

A is not, it's an arithmetic sequence where each term is +5 greater than the previous

B is because each term is 1/2 the previous term (common ratio of 1/2) thus making it a geometric sequence

C is not, it's the Fibonnaci sequence which is neither an arithmetic or geometric sequence

D is because each term is 1/3 the previous term (common ratio of 1/3) thus making it a geometric sequence

The NPV is positive, therefore, this order can be placed.

The calculation of probability of default will be:

= 1 / 200 = 0.005

We use NPV formula to find out whether the order should be filled or not

NPV = -Variable cost per unit +[(1-π)(price of equipment/1+percentage of return)]

= -$1.6 million +[(1-0.005)($1.725m/1.018)]

= -$1.6m + $1.6860

= 0.0860 million

As NPV is positive this order can be placed.

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Answer:

11 3/13% per annum

Step-by-step explanation:

to find rate= 100×simple interest ÷ (principal ×time)

simple interest=total amount - principal

=1,500,000-1,300,000

=200,000

=200,000 ×100÷ 1,300,000×4

=3 11/13% per annum

Answer:

d i did the thing

Step-by-step explanation:

The area of triangle is 16 square inches.

Area of triangle is half times the product of base and height.

i.e., A-= 1/2 *b *h

Given:

b= 8 inches, h= 4 inches

Now, Area of triangle

= 1/2*b*h

= 1/2 * 8 *4

=1/2*32

=16.

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It would take approximately 11 years for this population to reach 7300.

Based on the information provided, the future size of the population can be modelled by using an exponential function:

F = Pe^rt

Substituting the given parameters into the formula, we have;

7300 = 5000e^{0.035 × t}

7300/5000 = e^{0.035t}

1.46 = e^{0.035t}

Taking the ln of both sides, we have:

ln(1.46) = ln(e^{0.035t})

0.378 = 0.035t

t = 0.378/0.035

Time, t = 10.8 ≈ 11 years.

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The logarithmic form of √x = y is ㏒ₓy = 1/2

We have, logarithmic form as;

Given √x = y

Convert the square root to power of 1/2 as that is what it represents, we get

x^1/2 = y

Then compare with the universal logarithmic form, we have

㏒ₓy = 1/2

Thus, the logarithmic form of √x = y is ㏒ₓy = 1/2

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Answer:

n>2      (see below)

Step-by-step explanation:

Triangle side rule says    n+2   + 5   must be greater than 9

    n+2   +5  > 9

    n > 2      

      (but it must be less than 12   o/w   5 + 9 will not be greater than n+2)  

The inverse function of f(x) = 5x/3 + 5 is f-1(x) = 3x/5 - 3

The function is given as:

f(x) = 5x/3 + 5

Rewrite as:

y = 5x/3 + 5

Swap x and y

x = 5y/3 + 5

Subtract 5 from both sides

x - 5 = 5y/3

Multiply through by 3

3x - 15 = 5y

Divide through by 5

y = 3x/5 - 3

Express as inverse function

f-1(x) = 3x/5 - 3

Hence, the inverse function of f(x) = 5x/3 + 5 is f-1(x) = 3x/5 - 3

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Answer:

Step 2

Step-by-step explanation:

In step 1, Gary had the right idea of multiplying both sides by 4 to eliminate the denominator, but in step 2, Gary makes a mistake since e should equal 10 not 8. 2.5 * 4 = 10 not 8

Answer:

B) Step 2

Step-by-step explanation:

Gary's mistake occurs in Step 2. He was right to multiply both sides of the equation by 4 to find e, but his error is in the multiplication of 2.5*4. 2.5*4 = 10, not 8.

Answer:

The method of matrixes that would be used for this question is Gaussian Elimination.

Step-by-step explanation:

Gaussian Elimination is a method of solving linear equations by transforming them into upper triangular form. In this case, we want to solve for the original amount of money placed in each account, which can be represented by variables x, y, and z. We can set up the equations as follows:

4x + 5.5y + 6z = 1,300

x + y + z = 24,500

4x = y

We can then use Gaussian Elimination to solve for x, y, and z.

The augmented matrix of the system of equations is [tex]\left[\begin{array}{ccc}-2&2&-1\\-3&-1&1\\-1&3&-2\end{array}\right] \left[\begin{array}{c}13&7\\0\end{array}\right][/tex]

The system of equations is given as:

-2x + 2y - z = 13

-3x - y + z = 7

-x + 3y - 2z = 0

Remove the variables x, y and z

-2      2    -1  13

-3    -1        1    7

-1     3      -2     0

Express as matrix

[tex]\left[\begin{array}{ccc}-2&2&-1\\-3&-1&1\\-1&3&-2\end{array}\right] \left[\begin{array}{c}13&7\\0\end{array}\right][/tex]

Hence, the augmented matrix of the system of equations is [tex]\left[\begin{array}{ccc}-2&2&-1\\-3&-1&1\\-1&3&-2\end{array}\right] \left[\begin{array}{c}13&7\\0\end{array}\right][/tex]

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Complete question

Write the augmented matrix of the system of equations given below. (10 points)

-2x + 2y - z = 13

-3x - y + z = 7

-x + 3y - 2z = 0