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4 years ago

A store experienced a 14.2% fall in takings in February, compared to the previous month.

1 answer:

34kurt4 years ago

4 0

say the amount in jan is x we know if we subtract off 14.2% of x (from x) we get Feb's amount

so x - 0.142x = 742,513.20 (notice we change 14.2% to 14.2/100 -- that is what % is short for -- or ,as a decimal 0.142)

we have 1x - 0.142x or (1-0.142)x on the left side
0.858 x = 742513.20

divide both sides by 0.858 to "solve for x"

x= 865,400

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Assume that the heights of men are normally distributed with a mean of "71.3" inches and a standard deviation of 2.1 inches. If

Elza [17]

Answer:

0.0021 = 0.21% probability that they have a mean height greater than 72.3 inches.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Assume that the heights of men are normally distributed with a mean of "71.3" inches and a standard deviation of 2.1 inches.

This means that $\mu = 71.3, \sigma = 2.1$

Sample of 36:

This means that $n = 36, s = \frac{2.1}{\sqrt{36}} = 0.35$

Find the probability that they have a mean height greater than 72.3 inches.

This is 1 subtracted by the pvalue of Z when X = 72.3. So

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

By the Central Limit Theorem

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

$Z = \frac{72.3 - 71.3}{0.35}$

$Z = 2.86$

$Z = 2.86$ has a pvalue of 0.9979

1 - 0.9979 = 0.0021

0.0021 = 0.21% probability that they have a mean height greater than 72.3 inches.

7 0

3 years ago

This one pleaseeeeeeeeee helpppp

stiks02 [169]

Answer: x = 9

Remember: cross-multiply (product of the means equals the product of the extremes) to solve this proportion. It forms a 'cross'-- hence its name.

4 8 4 and 2 are the extremes
------ = ------ 8 and x - 8 are the means
x - 8 2

(4)(2) = 8(x - 8) Simplify
8 = 8(x - 8) Distributive Property
8 = 8x - 64 Add 64 to both sides
72 = 8x Divide by 8 for both sides
9 = x Answer!
x = 9 Answer!

7 0

3 years ago

Read 2 more answers

In the function y = -5x2 + 5 , the graph will open down, be shifted up 5 units, and it will be wider than the parent function y

masya89 [10]

We are considering the function $y=f(x)=-5 x^{2} +5$

the graph of this function is the graph of the parent function $y=g(x)= x^{2}$ after applying the following steps.

1. multiply by -1:

$-x^{2}$ is $x^{2}$ reflected with respect to the x-axis.

2. multiply by 5:

$-5x^{2}$ is $-x^{2}$ shrinked towards the y axis.

for example, consider x=1/2

$-5x^{2}=-5( \frac{1}{2} )^{2}=-5* \frac{1}{4}= \frac{-5}{4}=-1.25$

$-x^{2}=- ( \frac{1}{2} )^{2}=- \frac{1}{4}=-0.25$

now this means that $-5x^{2}$ is "lower" than $-x^{2}$ , which means there is a shrinking of the graph.

3. add 5:

$-5 x^{2} +5$ is $-5 x^{2}$ shifted 5 units up.

4.

so the graph of the function a) opens down, b) is shifted 5 units up, c) is not wider than the parent function

Answer: False

6 0

4 years ago

Ken can walk 40 dogs in 8 hours how many can ken walk in 12 hours

Jlenok [28]

It would half 8 hours to get 4 hours and 20 dogs
Next add the 4 hours and 20 dogs to the 8 hours and 40 dogs so
4+8=12 hours
20+40=60 dogs
So Ken can walk 60 dogs in 12 hours

6 0

3 years ago

I NEEDZ HELPZZZZZZZZ!!!! btw i didnt mean to click d)20..

joja [24]

The slope is 0 or answer B.)

8 0

3 years ago