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Jet001 [13]
3 years ago
6

Can someone please solve this for me

Mathematics
1 answer:
anzhelika [568]3 years ago
6 0

it equals 8700 aka c

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Please help, i’m not sure if i’m correct
hammer [34]

Answer:

Looks good to me! :)

Step-by-step explanation: Hope you have a great rest of your day!

7 0
3 years ago
What is –9.2(8x – 4) + 0.7(2 + 6.3x) simplified? A:–69.19x – 32.39 B:–69.19x + 38.2 C:–72.2x + 41.21 D:75x – 338.2
Veseljchak [2.6K]
-9.2(8x - 4) + 0.7(2 + 6.3x)....distribute thru the parenthesis
-73.6x + 36.8 + 1.4 + 4.41x ...combine like terms
(-73.6x + 4.41x) + (36.8 + 1.4) =
-69.19 + 38.2 <====

3 0
4 years ago
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Larry runs a dog-walking service. He charges $5/h plus a flat fee of $6. One day, he earned $16. Determine which
frosja888 [35]

Answer:

5h+6=16

Step-by-step explanation:

He gets 6 dollars and for every hour he walks he gets 5 dollars so he must of walked dogs for 2 hours and plug it in you get your equation.

4 0
3 years ago
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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
SOLVE THE EQUATION SHOW YOUR WORK 3x = 45
Hunter-Best [27]

Answer:

x = 15

Step-by-step explanation:

3x = 45

x = 45/3

<em><u>x = 15</u></em>

4 0
3 years ago
Read 2 more answers
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