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Elden [556K]
3 years ago
15

The probability of a getting a new sales contract with any given client is 0.29. A sales representative works with 18 clients th

is week. What is the expected number of contracts the sales representative will get this week?
Mathematics
1 answer:
Radda [10]3 years ago
7 0

Answer:

The expected number of contracts the sales representative will get this week is 5.22

Step-by-step explanation:

For each client, there are only two possible outcomes. Either there is a sales contract, or there is not. The probability of having a sale contract with each client is independent of other clients. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

In this problem, we have that:

n = 18, p = 0.29

What is the expected number of contracts the sales representative will get this week?

E(X) = np = 18*0.29 = 5.22

The expected number of contracts the sales representative will get this week is 5.22

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Jack is getting balloons for his uncle's birthday party. He wants each balloon string to be 6 feet long. At the party store, str
SCORPION-xisa [38]

Answer:

110 yards of string

Step-by-step explanation:

Converting the feet of the string of 1 balloon to yard

1 foot = 1/3 yard

6 feet = x

x = 6 × 1/3 yard

x = 2 yard

Hence 1 balloon uses 2 yard of string

Since:

1 balloon = 2 yard of string

55 ballons = x

x = 55 × 2 yard of string

x = 110 yards of string

Hence, for 55 balloons he would need 110 yards of string.

6 0
3 years ago
A rectangular lot that measures 150 ft by 200 ft is completely fenced. What is the approximate length, in feet, of the fence?
Ivenika [448]
300,000 because you have to just multiply 150 by 200
3 0
4 years ago
Jensen Tire & Auto is in the process of deciding whether to purchase a maintenance contract for its new computer wheel align
sasho [114]

Answer:

Step-by-step explanation:

Hello!

The given data corresponds to the variables

Y:  Annual Maintenance  Expense ($100s)

X: Weekly Usage  (hours)

n= 10

∑X= 253; ∑X²= 7347; \frac{}{X}= ∑X/n= 253/10= 25.3 Hours

∑Y= 346.50; ∑Y²= 13010.75; \frac{}{Y}= ∑Y/n= 346.50/10= 34.65 $100s

∑XY= 9668.5

a)

To estimate the slope and y-intercept you have to apply the following formulas:

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} } = \frac{9668.5-\frac{253*346.5}{10} }{7347-\frac{(253)^2}{10} }= 0.95

a= \frac{}{Y} -b\frac{}{X} = 34.65-0.95*25.3= 10.53

^Y= a + bX

^Y= 10.53 + 0.95X

b)

H₀: β = 0

H₁: β ≠ 0

α:0.05

F= \frac{MS_{Reg}}{MS_{Error}} ~~F_{Df_{Reg}; Df_{Error}}

F= 47.62

p-value: 0.0001

To decide using the p-value you have to compare it against the level of significance:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The decision is to reject the null hypothesis.

At a 5% significance level you can conclude that the average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

b= 0.95 $100s/hours is the variation of the estimated average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

a= 10.53 $ 100s is the value of the average annual maintenance expense of the computer wheel alignment and balancing machine when the weekly usage is zero.

c)

The value that determines the % of the variability of the dependent variable that is explained by the response variable is the coefficient of determination. You can calculate it manually using the formula:

R^2 = \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{[sumY^2-\frac{(sumY)^2}{n} ]} = \frac{0.95^2[7347-\frac{(253)^2}{10} ]}{[13010.75-\frac{(346.50)^2}{10} ]} = 0.86

This means that 86% of the variability of the annual maintenance expense of the computer wheel alignment and balancing machine is explained by the weekly usage under the estimated model ^Y= 10.53 + 0.95X

d)

Without usage, you'd expect the annual maintenance expense to be $1053

If used 100 hours weekly the expected maintenance expense will be 10.53+0.95*100= 105.53 $100s⇒ $10553

If used 1000 hours weekly the expected maintenance expense will be $96053

It is recommendable to purchase the contract only if the weekly usage of the computer is greater than 100 hours weekly.

4 0
3 years ago
G(x) = –(x + 3)2 – 4?
professor190 [17]

Answer:

g(x)=-x-5

Step-by-step explanation:

Simplify parenthesis by distribution:

g(x)=-x-3+2-4

Addition and Subtraction:

-3+2=-1, -1-4=-5

Finalize since no more simplification:

g(x)=-x-5

Pls give thx and brainliest wanna level up

6 0
3 years ago
Your are 5 feet tall. What is your height in meters?i
Svetach [21]

Answer: 5 ft = 1.5240 m

3 0
3 years ago
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