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ludmilkaskok [199]

2 years ago

9

Kenya applied for a bank loan. Kenya calculated he would pay $78.00 interest if the rate was 12%. Use the percent equation to fi

nd how much money Kenya wanted to borrow from the bank.
78 is 12% of what number?
A
$936.00

B
$93.60

C
$650.00

D
$1,200.00

2 answers:

Morgarella [4.7K]2 years ago

8 0

78 is 12% of what number

78 is 12% of c 650

AfilCa [17]2 years ago

5 0

78 is 12% of what number?
The answer is c. $650.00

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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,564 miles, with a standard

DerKrebs [107]

Answer:

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 48564, \sigma = 3293, n = 281, s = \frac{3293}{\sqrt{281}} = 196.44$

What is the probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct?

This is the pvalue of Z when X = 48101. So

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

By the Central Limit Theorem

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

$Z = \frac{48101 - 48564}{196.44}$

$Z = -2.36$

$Z = -2.36$ has a pvalue of 0.0091

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

6 0

2 years ago

5?10 5 5, dot, 10, start superscript, 5, end superscript is how many times as large as 1\cdot10^51?10 5 1, dot, 10, start supers

max2010maxim [7]

Answer:

5 times as large

Step-by-step explanation:

The ratio of the two numbers tells you how many times as large one is as the other:

$\dfrac{5\cdot 10^5}{1\cdot 10^5}=\dfrac{5}{1}\cdot\dfrac{10^5}{10^5}=5$

4 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

What is 1/2÷3 as a fraction?

dlinn [17]

Answer: 1/6

Step-by-step explanation:

1/2 divided by 3/1 turns into 1/2 x 1/3 and when multiplying numerators and denominators with each other, you get 1/6.

8 0

3 years ago

Read 2 more answers

a circle is centered on point b. points a, c and d lie on its circumference. if angle abc measures 40 degrees, what does angle a

grigory [225]

Answer:

320 degrees

Step-by-step explanation:

There are 360 degrees in a circle. 360-40=320

3 0

3 years ago