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timurjin [86]
3 years ago
5

The manager of a concert hall is keeping track of the number of people who pass through the doors before a concert begins. Using

his data, he writes a function showing the number of people who enter each minute. Which set of numbers represents the range of this function?
A. whole numbers greater than 0
B. real numbers greater than 0
C. all whole numbers
D. all real numbers
GIVING BRAINLIEST//20 points**
Mathematics
1 answer:
nignag [31]3 years ago
7 0

Answer:

i am pretty sure it's B. real numbers greater than 0

Step-by-step explanation:

please let me know if that is correct :D

if it is correct please give brainIest

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AlexFokin [52]

P=(L•2)+(W•2)

L=16 in and W=11 in

P=(16•2)+(11•2)

P=54

A=L•W

L=16 in and W=11 in

A=16•11

A=176

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Solve the following system of equations. -3x + 5y = 80 x + 5y = 40 A. x = -10, y = 10 B. x = 2, y = 17.2 C. x = 8, y = 6.4 D. x
hammer [34]
-3x+5y=80 if the x = -10 and y = 10
x+5y=40 if the x= -10 and y= 10
5 0
3 years ago
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and
masya89 [10]

Answer:

a) There is a 100% probability that the (sample) average time waiting in line for these customers is less than 10 minutes.

b) There is a 100% probability that the (sample) average time waiting in line for these customers is between 5 and 10 minutes.

c) There is a 0% probability that the (sample) average time waiting in line for these customers is less than 6 minutes.

d) Because there are less observations, it would be less accurate.

e) Because there are moreobservations, it would be more accurate.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. This means that \mu = 8.2, \sigma = 1.5.

Suppose that a random sample of n = 49 customers is observed

This means that s = \frac{1.5}{\sqrt{49}} = 0.21.

(a) Less than 10 minutes.

This probability is the pvalue of Z when X = 10. So:

Z = \frac{X - \mu}{s}

Z = \frac{10 - 8.2}{0.21}

Z = 8.57

Z = 8.57 has a pvalue of 1.

This means that there is a 100% probability that the (sample) average time waiting in line for these customers is less than 10 minutes.

(b) Between 5 and 10 minutes.

This probability is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 5.

From a), we have that the zscore of X = 10 has a pvalue of 1.

For X = 5.

Z = \frac{X - \mu}{s}

Z = \frac{5 - 8.2}{0.21}

Z = -15.24

Z = -15.24 has a pvalue of 0.

Subtracting, we have that there is a 100% probability that the (sample) average time waiting in line for these customers is between 5 and 10 minutes.

(c) Less than 6 minutes.

This probability is the pvalue of Z when X = 6. So:

Z = \frac{X - \mu}{s}

Z = \frac{6 - 8.2}{0.21}

Z = -10.48

Z = -10.48 has a pvalue of 0.

This means that there is a 0% probability that the (sample) average time waiting in line for these customers is less than 6 minutes.

(d) If you only had two observations instead of 49 observations, would you believe that your answers to parts (a), (b), and (c), are more accurate or less accurate? Why?

The less observations there are, the less acurrate our results are.

So, because there are less observations, it would be less accurate.

(e) If you had 1,000 observations instead of 49 observations, would you believe that your answers to parts (a), (b), and (c), are more accurate or less accurate? Why?

The more observations there are, the more acurrate our results are.

So, because there are moreobservations, it would be more accurate.

8 0
3 years ago
The employees of a hardware store ordered lunch from a local delicatessen. The lunch consisted of 4 turkey sandwiches and 7 orde
AleksandrR [38]

Answer:

Charges of Turkey is $6.25

Charges of French Fries is $1.90

Step-by-step explanation:

Given

Represent Turkeys with T and French Fries with F;

4T + 7F = 38.3

5T + 5F = 40.75

Required

Determine the values of T and F

5T + 5F = 40.75

Divide through by 5

T + F = 8.15

Make F the subject

F = 8.15 - T

Substitute 8.15 - T for F in the first equation

4T + 7F = 38.3

4T + 7(8.15 - T) = 38.30

4T + 57.05 - 7T = 38.30

Collect Like Terms

4T  - 7T = 38.30 - 57.05

-3T = -18.75

Solve for T

T = -18.75/-3

T = 6.25

Recall that:

F = 8.15 - T

F = 8.15 - 6.25

F = 1.90

6 0
3 years ago
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