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2 years ago

14

12. Ms. Hall wants to buy 2 pairs of shoes. It was buy one at original price of $70 and get one 25% off. How much will Ms. Hall

pay for her shoes?

1 answer:

kaheart [24]2 years ago

4 0

Answer:

17.50

Step-by-step explanation:

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One reading an Antarctic research station showed that the temperature was -35 degrees Celsius what is this temperature in degree

Musya8 [376]

It is very simple it would be positive but would remain 35

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3 years ago

.0096 divided by 8 I will give u 10 points

balu736 [363]

Answer:

0.0012

Step-by-step explanation:

0.0096/8=0.0012

8 0

3 years ago

Read 2 more answers

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

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:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

2 years ago

A man pays only $60 for an item which originally sells for $100. What percent of the original price does the man pay for the ite

Anni [7]

Answer: 40%. Alternatively, you can think about it this way. The item is 40% off. This means you'll pay 60.000% of the total cost (100% - 40% = 60.000%).

4 0

3 years ago

Read 2 more answers

A sample of salary offers (in thousands of dollars) given to management majors is: 48, 51, 46, 52, 47, 48, 47, 50, 51, and 59. U

balu736 [363]

Answer:

Step-by-step explanation:

number of samples, n = 10

Mean = (48 + 51 + 46 + 52 + 47 + 48 + 47 + 50 + 51 + 59)/10 = 49.9

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (48 - 49.9)^2 + (51 - 49.9)^2 + (46 - 49.9)^2+ (52 - 49.9)^2 + (47 - 49.9)^2 + (48 - 49.9)^2 + (47 - 49.9)^2 + (50 - 49.9)^2 + (51 - 49.9)^2 + (59- 49.9)^2 = 128.9

Standard deviation = √128.9/10 = 3.59

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 10 - 1 = 9

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.262

Margin of error = 2.262 × 3.59/√10

= 2.57

the lower limit of this confidence interval is

49.9 - 2.57 = 47.33

the lower limit of this confidence interval is

49.9 + 2.57 = 52.47

So it is false

6 0

2 years ago