What is r = wp, for p

You are purchasing books summer reading for $4.50 each. Part A: Your total cost, C, depends on,The input is,and the output is,Pa

Crazy boy [7]

Answer:

Part A) see the explanation

Part B) $C(12)=\$54$

Step-by-step explanation:

Part A)

we know that

A relationship between two variables, x, and C, represent a proportional variation if it can be expressed in the form $k=\frac{C}{x}$ or $C=kx$

Let

C ---> is the total cost (represent the output or dependent variable)

x ----> is the number of books (represent the input or independent variable)

In this problem

we have a proportional relationship between the variables x and C

Part B) we know that

The linear equation is given by

$C=kx$

The constant of proportionality k is the same that the slope or unit rate of the linear equation

In this problem the constant k is given

$k=\$4.50\ per\ book$

so

$C=4.50x$

For x=12 books

substitute in the equation and solve for C

$C=4.50(12)=\$54$

In function notation

$C(12)=\$54$

5 0

2 years ago

Consider the following sample of observations on coating thickness for low-viscosity paint.

Julli [10]

Answer:

a) $\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

b) Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

c) $P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

d) $Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

Step-by-step explanation:

Data given:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Part a

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

Part b

For this case in order to calculate the median we need to put the data on increasing way like this:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

Part c

For this case we can assume that the mean is $\mu = 1.3538$

And we can calculate the population deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}}$

And if we replace we got: $\sigma= 0.3105$

And assuming normal distribution we have this:

$X \sim N (\mu = 1.3538, \sigma= 0.3105)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

Part d

The median is defined as :

$Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

6 0

2 years ago

Consider a prolific breed of rabbits whose birth and death rates, β and δ, are each proportional to the rabbit population P = P(

KonstantinChe [14]

Answer:

(P(t)) = P₀/(1 - P₀(kt)) was proved below.

Step-by-step explanation:

From the question, since β and δ are both proportional to P, we can deduce the following equation ;

dP/dt = k(M-P)P

dP/dt = (P^(2))(A-B)

If k = (A-B);

dP/dt = (P^(2))k

Thus, we obtain;

dP/(P^(2)) = k dt

((P(t), P₀)∫)dS/(S^(2)) = k∫dt

Thus; [(-1)/P(t)] + (1/P₀) = kt

Simplifying,

1/(P(t)) = (1/P₀) - kt

Multiply each term by (P(t)) to get ;

1 = (P(t))/P₀) - (P(t))(kt)

Multiply each term by (P₀) to give ;

P₀ = (P(t))[1 - P₀(kt)]

Divide both sides by (1-kt),

Thus; (P(t)) = P₀/(1 - P₀(kt))

4 0

2 years ago

A company buys paper at the rate of $10.50 per package for the first 10 packages, $8.50 per package for the next 10 packages, an

dexar [7]

Answer:

25?

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

8th grade math fsa countdown answers

Liula [17]

https://jhmsmath.weebly.com/uploads/2/7/7/5/27756807/8th_grade_countdown.pdf go to this link page 52-54

6 0

2 years ago