Hi i need help with this question.

(a) There are 10 units must be sold to maximize profit.

(b) The maximum profit is 480.

what is profit?

A profit is an income that is given to the owner in an efficient market production process in accounting. The owner's primary interest in the income-formation process of market production is profit, which is a measure of profitability. Different profit metrics are frequently employed.

Given:

A company sales x units of a product, its profits P can be determined

using the profit function P(x) = -2x² + 40x + 280       ..(1)

We have to find the maximum profit.

Differentiate equation (1) w.r.t x

     P'(x) = -4x +40            ..(2)

-4x + 40 = 0

        4x = 40

          x = 10

Again differentiate equation (2) w.r.t x

P''(x) = -4 which is negative.

Hence, there are 10 units must be sold to maximize profit.

Now, to find the maximum profit.

Plug x = 10 in equation (1)

P(10) = -2(10)^2 + 40(10) + 280

        = -200 + 400 + 280

P(10) = 480

Hence, the maximum profit is 480.

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Answer: 35/10

Step-by-step explanation:

if you have 7/2 and need a denominator of 10 all you need to do is multiply 7/2 by 5/5.

7  by  5     35

2  by  5     10

The value of 7/2 with a denominator of 10 is 35/10

Suppose the given fraction is [tex]\dfrac{a}{b+c}[/tex]

Then the conjugate of the denominator is given by b - c

Thus, rationalizing the fraction will give us

[tex]\dfrac{a}{b+c} \times \dfrac{b-c}{b-c} = \dfrac{a(b-c)}{b^2 - c^2}[/tex]

(we used the identity [tex](b-c)(b+c) = b^2 - c^2 )[/tex]

We actually rationalize just for the use of that later denominator or numerator

We have 7/2 and need a denominator of 10 all you need to do is multiply 7/2 by 5/5.

7 × 5 = 35

2 × 5 = 10

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Answer:

[tex]\pi[/tex]

Step-by-step explanation:

Area = [tex]\pi x^{2}[/tex]  For a full circle.  This is 1/4 of a circle

A= [tex]\frac{\pi 2^{2} }{4}[/tex] = [tex]\frac{\pi 4}{4}[/tex] = [tex]\pi[/tex]

Answer : The area of the shape is, 111.02 units 2

a) X1+X2 have distribution Bi(n1+n2, p)

b) P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).

b)

Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).

If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹

If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹

Thus,

P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

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The exponential function Q=Q0 (1+r) t can be used to describe exponential growth or decay depending on the value of r. If r is less than 0, the function can be used to describe exponential growth, whereas if r is greater than 0, the function can be used to describe exponential decay.

The exponential function Q=Q0 (1+r) t can be used to describe exponential growth or decay. The function is determined by three variables: Q0, which is the starting value of the function; r, which is the growth or decay rate of the function; and t, which is the time when the function occurs. If r is less than 0, then the function can be used to describe exponential growth, meaning that the output of the function will increase over time. On the other hand, if r is greater than 0, then the function can be used to describe exponential decay, meaning that the output of the function will decrease over time. The exponential function can be used to describe a variety of different processes, such as population growth, compound interest, and radioactive decay. It can also be used to predict future values based on past values by using the growth or decay rate. Thus, the exponential function is a powerful tool for describing and predicting exponential growth and decay.

Q=Q0 (1+r) t

For example, if Q0 = 10, r = 0.5, and t = 5, then

Q=10 (1+0.5)5

Q= 10 x 1.53

Q= 15.3

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Answer:

$3.6 saved

Step-by-step explanation:

Shop A: -

7 cookies    ---->   $5.60

24 cookies   ----->  $x

x = (24 x 5.60) / 7

x = $19.2

Shop B: -

5 cookies  ---->  $4.75

24 cookies  ---- > $y

y = (24 x 4.75) / 7

y = $22.8

Saved = Shop B - Shop A

Saved = 22.8 - 19.2

Saved = $3.6

I hope my answer helps you.

The interval for horizontal distance is

(59.9 ,74.2)  ( ranging from 59.9 feet to 74.2 feet)

As per given in the question,

The length of the ladder (l) is 40feet

The angle of elevation in the triangle is [tex]60^0[/tex]

a) In a Δ ABC

One angle is given as 60 and the one side length (length of ladder) is given as 40 feet

The height of the slide is

[tex]sin\theta=\frac{opposite}{hypotenuse}[/tex]

[tex]sin60^0[/tex] = [tex]\frac{h}{40}[/tex]                                 where the value of   ( [tex]sin60^0={\frac{\sqrt{3} }{2} }[/tex])

=> [tex]\frac{\sqrt{3} }{2}[/tex] = [tex]\frac{h}{40}[/tex]

=> h = 20[tex]\sqrt{3}[/tex]

Approximately the height of the slide is 34.64 feet

b)    In Δ DEF,

        [tex]\theta = tan^{-1} \frac{h}{d}[/tex]                                                           ----( 1)

=>  [tex]Tan^{-1} (\frac{34.6}{d} )[/tex]

c)  the angle of depression is [tex]\theta=25^0[/tex] and [tex]\theta=30^0[/tex]

first for the value [tex]\theta=30^0[/tex]

substitute the value of [tex]\theta[/tex] in equation (1)

[tex]tan30^0=(\frac{34.63)}{d})[/tex]            where the value of ( [tex]Tan 30^0=\frac{1}{\sqrt{3} }[/tex])

=> [tex]\frac{1}{\sqrt{3} } =\frac{34.63}{d}[/tex]

=> d = 59.99

now

substitute the value [tex]\theta=25^0[/tex] in equation (1)

[tex]tan25^0=(\frac{34.6}{d} )[/tex]

=> d = 74.28

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The unit rates for each animal in mass per heart rate (beats/minute) are as follows:

Cat               13.33g/1 beat per min

Cow           12,307.69g/1 beat per min

Hamster       1.36g/1 beat per min

Horse          2,666.67g/1 beat per min

The unit rate is the ratio of two different variables expressed in a common denominator.

The unit rate is the quotient of the division operation between, for instance, the mass of an animal and its heart rate.

In this instance, the mass of the animal is the numerator with the heart rate used as the denominator.

Animal      Heart Rate     Mass (g)        Unit Rate

                (beats/min)                      (mass per heart rate)

Cat                150            2,000             13.33g/1 beat per min (2,000/150)

Cow               65         800,000    12,307.69g/1 beat per min (800,000/65)

Hamster        44                60                 1.36g/1 beat per min (60/44)

Horse          450      1,200,000   2,666.67g/1 beat per min (1,200,000/450)

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There is a 32.81% chance (0.3182) that there will be between 32647 and 43559 ants on a randomly chosen acre.

The Z-score calculates how far a measure deviates from the mean by standard deviation. We look at the p-value linked with this Z-score after determining the Z-score by consulting the z-score table. This p-value represents the likelihood that the measure's value is less than X, or the percentile of X. We may calculate the likelihood that the value of the measure exceeds X by subtracting 1 from the pvalue.

The zscore of a measure X in a set with mean and standard deviation may be calculated as follows:

Z= X-μ/σ

μ = 42000   σ= 12275

The pvalue of Z when X = 32647 is deducted from the pvalue of Z when X = 43559 to get this value. So

X = 43559

z= 0.13 has a pvalue of 0.5517

X = 32647

z=  -0.76 has a pvalue of 0.2236

0.5517 - 0.2236 = 0.3281

0.3182 = 32.81% probability that a randomly selected acre has between 32647 and 43559 ants.

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The equivalent expression for the given expression is [tex]x^{\frac{5}{3} }y^{\frac{1}{3} }[/tex]. Therefore, option B is the correct answer.

Equivalent expressions are expressions that work the same even though they look different. If two algebraic expressions are equivalent, then the two expressions have the same value when we plug in the same value for the variable.

The given expression ∛x⁵y.

expression is equivalent to ∛x⁵y is

[tex](x^5y)^{\frac{1}{3} }[/tex]

= [tex]x^{\frac{5}{3} }y^{\frac{1}{3} }[/tex]

The equivalent expression is [tex]x^{\frac{5}{3} }y^{\frac{1}{3} }[/tex]. Therefore, option B is the correct answer.

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There are 1307674368000 possible methods to remove each ball from the box one at a time.

How likely something is to occur is known as its probability. We can discuss the probabilities of various outcomes—how likely they are—when we aren't sure how a particular event will turn out.  examination of events subject to probability.

given,

the quantity of balls inside a box ,

four red balls

6 balls in blue

four green balls

The variety of ways the balls can be taken out of the box one at a time must be determined.

Here,

Combination is possible because order is imperceptible;

mixture formula; nCr is equal to n!/r1(n-r)!

Here, There are 15 balls in the box overall (5 + 4 + 6 balls).

Then.

¹⁵C₁₅ = 15! / 15! (15 - 15)! = 15! = 1307674368000

that is

There are 1307674368000 ways in which we can take the balls out of the box one at a time.

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63 mg of anesthetic is present in the patient’s circulatory system.

The operation can last for 20 minutes.

During surgery, a patient’s circulatory system requires at least 50 mg of an anesthetic. The amount of anesthetic present t hours after 70 mg of anesthetic is administered is given by the following,

T(t) = 70 × (0.727) ^t

Time = 1/3 hours

T (1/3) = 70 × (0.727) ^1/3

            = 70 × ∛ 0.727

             = 70 × 0.899176

            = 62.94232

To the nearest mg: 63mg

Now,

63 = 70 × (0.727) ^t

63/70 = (0.727) ^t

0.9 = (0.727) ^t

Log0.9 = log (0.727) ^t

Log0.9 = t × log (0.727)  

T = log0.9 / log0.727

T = 0.330461

T = 0.330461 hours

  = 20 minutes

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Answer:

The division algorithm is an algorithm that specifies how to perform the division of one number (called the dividend) by another number (called the divisor) to obtain the quotient and the remainder.

For example, if we want to divide 15 by 4, we can use the division algorithm to find the quotient and remainder. In this case, we have:

15 = 4 * 3 + 3

So, the quotient is 3 and the remainder is 3.

If it's not clear for you yet, you can wait for another answer, maybe better explained.

Easier example (the classic one with apples):

Imagine you have a bunch of apples, and you want to divide them equally among a group of friends. The division algorithm can help you figure out how many apples each friend gets, and how many apples are left over.

Here's an example, let's say you have 15 apples, and you want to divide them among 4 friends. Using the division algorithm, we can figure out that each friend gets 3 apples, and there will be 3 apples left over.

To use the division algorithm, you start by dividing the number of apples (15) by the number of friends (4). This gives us the number of apples each friend gets, which is called the "quotient." In this case, the quotient is 3.

Then, we take the remainder, which is the number of apples that are left over after we divide them evenly among the friends. In this case, the remainder is 3.

So, each friend gets 3 apples, and there are 3 apples left over.

For two events M and N, P(M)=0.6, P(N|M)=0.9, and P(N|M′)=0.2. Then P(M′|N.) is 0.7

This can be easily solved if we know the formula:

P(A or B) = P(A) + P(B) - P(A and B)

So, for this problem, we can modify the formula to be:

P(M or N) = P(M) + P(N) - P(M and N)

Using given information, we can solve:

P(M or N) = P(M) + P(N) - P(M and N)

P(M or N) = 0.6 + 0.2 - 0.1

P(M or N) = 0.7

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The weighted average of the average weights of each individual subgroup, where the weights represent the population proportions in each subgroup, provides the average weight of the population's constituents.

This can be mathematically represented as: average population weight = _(i=1) r P i * w i, where P i is the population's share of subgroup I and w i is the weight of the subgroup's average members.

For instance, if the population is divided into three subgroups with proportions of P1 = 0.2, P2 = 0.5, and P3 = 0.3, and the subgroups' average weights are P1 = 150 lbs, P2 = 200 lbs, and P3 = 175 lbs, respectively, then the population's average weight is P1 * 150 lbs 180 lbs is equal to +0.5 * 200 lbs + 0.3 * 175 lbs.

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The number of solution to the equation is 210 when consider the number 22 as being composed of 22 distinct items, with the variables x₁, x₂ and x₃.

Given that,

Consider the number 22 as being composed of 22 distinct items, with the variables x₁, x₂ and x₃ . acting as the three categories into which these items are sorted. Since each x is a true affirmative, your response is accurate. Put one item in each category to begin with, then divide the remaining items across the categories. The number of classification schemes for all the objects is the number of solutions to the equation satisfying the given criteria. In light of this, your response of 20 is erroneous.

We know that,

Number 22 is divided into 22 individual items ,

The three categories in which these items are placed are x₁, x₂ and x₃ .

x₁+x₂+x₃=22----->equation(1)

(x₁-1)+(x₂-1)+(x₃-1)=22-3=19

Let

a=x₁-1,

b=x₂-1 and

c=x₃-1.

So,

a+b+c= 19 ----->equation(2)

So , the number of ways of arranging 21 characters in which 19 are same(A) and 2 are same (J)

= 21!/(2! × 19!)

= 21×10 = 210

Therefore, the number of solution to the equation is 210 when consider the number 22 as being composed of 22 distinct items, with the variables x₁, x₂ and x₃.

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449.86

To compute the moment of the second sample, we need to compute the sum of the squares of the sample values.

This formula is:

Sum = ∑(x_i^2)

After inserting the example values:

Sum = 6.2^2 + 7.4^2 + 9.6^2 + 11.5^2 + 13.5^2

Simplifying this formula to:

Total = 38.44 + 54.76 + 92.16 + 132.25 + 181.25

Adding these terms gives us the following final result:

Total = 449.86

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P = 1 / n (Celia is first and Felicity is last in line) (n - 1)

P (in the line, Celia and Felicity are next to one another) = 2/n

A class consists of n kids.

These n kids line up for recess.

Out of n kids, there are two kids named Celia and Felicity.

(b) There are n positions and n! ways for n youngsters to stand in a line.

Consequently, n is the entire number of possible situations!

Felicity is standing in the last position while Celia is in the first (fixed position)

The remaining (n - 2) children can stand in the remaining 8 (n - 2) positions! ways

As a result, Celia is first in line and Felicity is last in line for the number of favorable situations, which is (n - 2)!

P (Celia is first in line and Felicity is last in line) = [(n - 2)! / n!]

P = (n - 2)! / [n(n - 1)(n - 2)!]

P = 1 / n(n - 1)

The likelihood that Celia and Felicity are on the same row as one another.

Consider Celia and Felicity as a single student, then multiply by the number of pupils (n - 1)

One can arrange (n - 1) pupils in (n - 1) positions! ways

However, there are two methods to organize Celia and Felicity among themselves.

There are therefore 2 good reasons to put Celia and Felicity adjacent to each other (n - 1)!

P(Celia and Felicity in the line are next to one another) = 2! * (n - 1)! / n!

P = 2 * (n - 1)! / n (n - 1)

P = 2/n

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After solving the equation by Completing the square , the answer is x = 0.4 , x = -8.4  .

In the question ,

it is given that ,

the equation is x² + 10x + 2 = 2x + 5,

rewriting the equation ,

we get ,

x² + 8x - 3 = 0

adding 3 to both sides ,

we get ,

x² + 8x = 3 ,

For the method of completing the square ,

we take half of the second term's coefficient, and square it, then add it to both sides.

that is 8/2 = 4 .

the equation becomes ;

x² + 2*x*4 + 4² = 3 + 4²

(x + 4)² = 3 + 16

(x + 4)² = 19

taking root on both the sides ,

we get ,

x + 4 = ±√19

So , x = -4 + √19    or   x = -4 - √19

Simplifying further ,

we get ,

x = 0.35889   or x = -8.35889

rounding to the nearest tenth we get

x = 0.4   or x = -8.4  .

Therefore , the solution of the given equation is x = 0.4 , x = -8.4  .

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Answer:

7 13/21 months

Step-by-step explanation:

10 2/3 ÷ 1 2/5 =

= 32/3 ÷ 7/5

= 32/3 × 5/7

= 160/21

= 7 13/21

Answer: 7 13/21 months

The maximum area of rectangle is 24, that can be calculated using the derivatives.

The area of rectangle = length * breadth,

The length of the rectangle is 2x units and the breadth is 3x+12 units,

So, A(x)=2x*(-3x+12)

=2x(-3x+12)

=-6x^2+24x

The derivative A'(x)=0

-12x+24=0

x=+2

Therefore, x=+2 is the critical point.

The second derivative.

The 2d derivative measures the immediately rate of alternate of the 1st derivative. The signal of the second derivative tells us whether or not the slope of the tangent line to is growing or decreasing.

A"(x)=-12<0

By the second derivative test, A(x) has maximum value at x=+2

Thus, the maximum are is

A(x)=2x(-3x+12)

A(2)=2(2)(-3(2)+12)

=4(6)

=2

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The equation in x and y for the line tangent to the curve is x - √3y - √3 = 0

A given curve can be defined by both its cartesian coordinates that is in x-y  form or in polar coordinates that is in terms of r and θ there exists a relationship between two sets of coordinates

x = r cosθ

y = r sinθ

The tangent to any curve f(x) is defined by f'(x) . In order to find tangent to a given terms of r and θ , we can calculate from the above to expressions f'(x)

f'(x) = dy/dx = dx/dθ ÷ dy/dθ

It is given in the question that

r = 2 sinθ

Replacing the value of r to the standard equation,

x = (2 sinθ )cosθ = sin2θ

At θ = π/3 , x = √3 / 2

y = (2 sinθ)sinθ = 2sin²θ

At θ = π/3 , y = 3/2

Differentiating both w.r.t θ

=> dx / dθ = 2cosθ

=> dy/dθ = 4sinθcosθ

So, dy/dx = 2cosθ / 4sinθcosθ

=> 1 / 2sinθ

Therefore , slope of tangent : m = 1/2sinθ ( θ = π/3)

=> m = 1 / 2sin(π/3)

=> m = 2 / 2√3

=> m = 1 /√3

Hence , the equation of line is

( y - 3/2 ) = 1 /√3 (x -  √3 / 2)

=> √3y - 3√3/2 = x - √3/2

=> x - √3y - √3 = 0

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(a) The standard deviation of the large wild cats is 25.5 (b)  The standard deviation of the birds in flight is 33 (c) The standard deviations of the two data sets show that the fastest recorded speeds of large wild cats deviate  more from the mean than the fastest recorded speeds of birds in flight . Option A

We should understand that standard deviation looks at the amount of deviation from the means of the set of data.

(a) (70+45+25+45+35+35+40+40+25)÷9 = 40

mean is = 40

(x-m) =30+5+-15+5+-5+-5+0+0+-15)

(x-m)²=(900+25+225+25+25+25+0+0+225)=1450/9

Standard deviation = √650

Standard deviation = 25.5

(b) x= (216+103+96+56+63+35+50+32+55+25+25+20)=776

Mean(m)=776/12=65

(x-m)=151+38+31+-9+-2+-30+-15+-33+-10+-40+-40+-45

Now the square of the deviations from the mean is

(x-m)²=(7396+1444+961+81+4+900+225+1089+100+1600+1600+2025)÷12

=1452

Standard deviation = √1452=33

(c) Therefore, from the calculations, the standard deviations of the two data sets show that the fastest recorded speeds of large wild cats deviate more from the mean than the fastest recorded speeds of birds in flight

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First possibility:  2 positive roots  0 negative roots  4 imaginary roots

Second possibility:  0 positive   0 negative   6 imaginary

Let f(x) = 4x6-4x3+8

f(x) has 2 sign changes.  So, by Descartes's Rule of Signs, f(x) has either                                                    2 positive roots or none.

f(-x) = 4(-x)6-4(-x)3+8 = 4x6+4x3+8

f(-x) has no significant changes.  So, by Descartes's Rule of Signs, f(x) has no negative roots.

f(x) has degree 6.  So, by counting multiplicities, there are a total of 6                                                                                               roots.

First possibility:  2 positive roots   0 negative roots  4 imaginary roots

Second possibility:  0 positive   0 negative   6 imaginary

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Out of the given choices, we choose 'not enough information given to answer this question'.

Since, we are given a p-value associated with a test of the slope coefficient equal to 0.028 which leads to 3% level of significance, it means that there is 3% chance of finding difference which is larger than the one given in our study given that the null hypothesis is true. Since, it is less than 5% level of significance, the null hypothesis is rejected and accept an alternative hypothesis. As this hypothesis is not given, we conclude that there is not enough information to give an answer to the question.

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The distance between Little Gold town and Old Gold town is 805km-107km.

What is distance?

The measurement of distance between two objects or points can be quantitative or occasionally qualitative. Distance in physics or common usage can refer to a physical length or an estimation based on other factors.

Distance is the sum of an object's movements, regardless of direction. Distance can be defined as the amount of space an object has covered, regardless of its starting or ending point.

Given: 107 km Big Gold town. 805km Little Gold town. 107km Old Gold town.

We have to find the distance between little gold town and old gold town.

Here, 805km Little Gold town. 107km Old Gold.

Hence, the distance between Little Gold town and Old Gold town is 805km-107km.

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Answer:  15,507.44.

Step-by-step explanation: To find the original price of the home before the down payment was made, you need to use the following formula:

Original price = Mortgage amount / (1 - Down payment percentage)

Plugging in the values given in the question, you get:

Original price = $170,000 / (1 - 8.6%)

The original price of the home before the down payment was made is $<<170000/(1-.086)=185507.44>>185,507.44.

This means that the down payment amount was $185,507.44 - $170,000 = 185507.44-170000=15,507.44.

When t = 3869, the rate of change of p is proportional to p.

A differential equation in mathematics is an equation that connects the derivatives of one or more unknown functions.

Applications often involve functions that reflect physical quantities, derivatives that depict the rates at which those values change, and a differential equation that establishes a connection between the three. Due to the prevalence of these relationships, differential equations are widely used in many fields, including engineering, physics, economics, and biology.

The primary focus of studying differential equations is on the solutions—the collection of functions that satisfy each equation—as well as the characteristics of those solutions.

Only the most basic differential equations can be solved using explicit formulas, but many features of a particular differential equation's solutions can be deduced without doing an exact calculation.

When a closed-form expression is used, it is common for.

According to our question-

The revised equation is then

P=5000exp(-0.0513t)

Find P at t=5 now.

P=5000 exp(-0.0513×5)

P=5000 ×exp(-0.25647) (-0.25647)

P=5000 × 0.77378

P=3868.89.

P = 3869. Roughly

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a. The computed regression equation's regression coefficients, which hold the other independent variable constant, show the impact of each independent variable on the dependent variable. Y will grow by 0.5906 units for every unit increase in x1 and by 0.498 units for every unit increase in x2, respectively.

b. The regression equation can be changed to predict y for the cases when x1 and x2 are 180 and 310, respectively: y = 29.1270 + 0.5906 * 180 + 0.4980 * 310 = 29.1270 + 105.208 + 156.38 = 290.82

c. The sum of squares resulting from regression (SSR) is denoted by the formula: SSR = _(i=1)n (y i - y)2.

SSR = 6,216.375.

d. R2 = SSR / SST is how it is calculated.

R2 = 6,216.375 / 6,724.125 = 0.924 e is the result of substituting the supplied values. A measurement of the overall significance of the link between the variables is the F statistic. The calculation is:

f. The following equation yields the mean square due to regression (MSR):

MSR is calculated as SSR / (r - 1), where r is the total number of independent variables. MSR = 6,216.375 / (2 - 1) = 6,216.375 / 1 = 6,216.375 is the result of substituting the provided numbers.

g. MSE = SSE / (n - r - 1), where n is the number of observations and r is the number of independent variables, is the formula for the mean square due to error (MSE).

MSE = 507.75 / (10 - 2 - 1) = 507.75 / 7 = 72.53. With (r - 1) and (n - r - 1) degrees of freedom, we must compare the derived F statistic to the crucial value of F at alpha = 0.05 in order to utilize the F test to assess the overall significance of the association between the variables at a significance level of alpha = 0.05.

The crucial value can be found via of F, we can employ a critical value table or a statistical software program. The critical value of F is roughly 3.35 when alpha = 0.05 and the degrees of freedom are (r - 1) = (2 - 1) = 1 in the numerator and (n - r - 1) = (10 - 2 - 1) = 7 in the denominator.

The numbers for MSR and MSE can be substituted to get the following result: F = MSR/MSE = 6,216.375 / 72.53 = 85.24

The estimated F statistic (85.24) is higher than the threshold value of 3.35, which allows us to reject the null hypothesis that there is no association between the variables at alpha = 0.05. This shows a meaningful relationship between the dependent variable and at least one of the independent variables.

i. We must compare the t statistic for x1 to the crucial value of t at alpha = 0.05 with (n - r - 1) degrees of freedom in order to conduct a t test for the significance of x1 at an alpha = 0.05 significance level. As stated by the formula: t = b1 / sb1

where b1 is the regression coefficient for x1 and sb1 is the coefficient's standard error. T = 0.5906 / 0.0813 = 7.26 is the result of substituting the supplied values.

Statistical software or a table of critical values can both be used to determine the critical value of t. The critical value of t at alpha = 0.05 with (n - r - 1) = (10 - 2 - 1) = 7 degrees of freedom is roughly 2.365.

The estimated t statistic (7.26) exceeds the critical value of 2.365, hence the null hypothesis that the coefficient for x1 is not substantially different from zero at alpha = 0.05 can be rejected.

j. To run a t test to determine whether x2 is significant atWith (n - r - 1) degrees of freedom and a significance level of alpha = 0.05, we must compare the t statistic for x2 to the critical value of t. The formula for the t statistic for x2 is: t = b2 / sb2, where b2 is the x2 regression coefficient and sb2 is the coefficient's standard error. When the specified values are substituted, the following results are produced:

t = 0.4980 / 0.0567 = 8.79

Statistical software or a table of critical values can both be used to determine the critical value of t. The critical value of t at alpha = 0.05 with (n - r - 1) = (10 - 2 - 1) = 7 degrees of freedom is roughly 2.365.

As a result of the calculated t statistic (8.79) exceeding the crucial value of 2.365, we

can disprove the null hypothesis that the coefficient for x2 at alpha = 0.05 is not statistically different from zero.

k. The t statistic and the coefficient's standard error can be used to create a 95% confidence interval for the coefficient of x2. B2 +/- t * sb2 yields the confidence interval.

The values are replaced to yield: 0.4980 +/- 2.365 * 0.0567 = (0.3778, 0.6182)

This suggests that we have a 95% confidence that the true value of the x2 coefficient falls inside the given range (0.3778, 0.6182).

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Answer:it isA

Step-by-step explanation:

Answer:

8x^6

Step-by-step explanation:

Add the exponents and multiply the numbers