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Because Helen was prepared to pay cash for a yellow car with a marked price of $9200 she was given a discount of $1012 what perc

Liula [17]

Answer:

11%

Step-by-step explanation:

The percentage of the discount is 1012/9200 x 100 = 11%

3 0

3 years ago

For this problem what I did was add all the measurements and I got 48 m. However, it is wrong. How do I go about solving the per

erma4kov [3.2K]

9514 1404 393

Answer:

66 m

Step-by-step explanation:

The perimeter is the sum of the measures of <em>all</em> of the sides. There are two side measures that are missing from the diagram.

The missing horizontal measure is ...

17 m - 8 m = 9 m

The missing vertical measure is ...

16m -7 m = 9 m.

If you add these to the sum you already calculated, you will get the correct answer:

48 m + 9 m + 9 m = 66 m . . . perimeter of the figure

_____

If you're paying attention, you see that the sum of the measures of the two shorter horizontal segments is the same as the measure of the longer horizontal segment. Likewise, the sum of the measurements of the two shorter vertical segments is the same as that of the longer vertical segment.

In other words, the perimeter of this (and any) L-shaped figure is the same as the perimeter of a rectangle having the same horizontal and vertical dimensions as the long sides of the figure.

P = 2(17 m +16 m) = 2(33 m) = 66 m

8 0

3 years ago

A stack of bricks has a

Helga [31]

Answer: 456 inches cubed

Step-by-step explanation:

Volume is the area of the base multiplied by the height so you have to do 38 times 12 to get 456.

4 0

3 years ago

LAST ATTEMPT MARKING AS BRAINLIEST!! (Graph the image of the figure using the dilation given)

Iteru [2.4K]

L’ (4.5,0) J’(-4.5, -4.5) K’(-4.5,4.5) i think

5 0

2 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

3 years ago