Ask question

Ask question

All categories

3 years ago

13

If you received a 5% raise and your final paycheck was 100(which includes your 5% raise), how much was your original paycheck?

1 answer:

Schach [20]3 years ago

8 0

so you were earning say "x", so "x" is the 100% of your paycheck.

but you're da bomb and so you got a raise of 5%, so the new amount of your paycheck is 100% + 5%, so is 105% really, and we happen to know that is $100.

$\bf \begin{array}{ccll} \$&\%\\ \cline{1-2} 100&105\\ x&100 \end{array}\implies \cfrac{100}{x}=\cfrac{105}{100}\implies 10000=105x \\\\\\ \cfrac{10000}{105}=x\implies 95.238\approx x$

You might be interested in

Solve the system of equations.

Semenov [28]

I hope this helps you

4 0

3 years ago

You roll 2 dice what is the probability that the sum of the dice is greater than 8 and that one die shows a six

MAXImum [283]

It'll be rare to do it

7 0

3 years ago

André solves the following system of equations by elimination.

Vaselesa [24]

Although the question is incomplete, I have seen it before.

The equations are:
3a + 6b = 12
-3a + 6b = -12

Adding the equations, we get:

12b = 0
And b = 0

8 0

3 years ago

Read 2 more answers

3. The adult men of the Dinaric Alps have the highest average height of all regions. The

ahrayia [7]

Using the normal distribution, it is found that:

3 - a) The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

3 - b) The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

4 - a) The 25th percentile for the math scores was of 71.6 inches.

4 - b) The 75th percentile for the math scores was of 78.4 inches.

<h3>Normal Probability Distribution </h3>

In a <em>normal distribution </em>with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

Question 3:

The mean is of 73 inches, hence $\mu = 73$ .

The standard deviation is of 3 inches, hence $\sigma = 3$ .

Item a:

The 40th percentile is X when Z has a p-value of 0.4, so <u>X when Z = -0.253</u>.

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

$-0.253 = \frac{X - 73}{3}$

$X - 73 = -0.253(3)$

$X = 72.2$

The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

Item b:

The minimum height is the 100 - 10 = 90th percentile is X when Z has a p-value of 0.9, so <u>X when Z = 1.28</u>.

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

$1.28 = \frac{X - 73}{3}$

$X - 73 = 1.28(3)$

$X = 76.84$

The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

Question 4:

The mean score is of 75, hence $\mu = 75$ .

The standard deviation is of 5, hence $\sigma = 5$ .

Item a:

The 25th percentile is X when Z has a p-value of 0.25, so <u>X when Z = -0.675</u>.

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

$-0.675 = \frac{X - 75}{5}$

$X - 75 = -0.675(5)$

$X = 71.6$

The 25th percentile for the math scores was of 71.6 inches.

Item b:

The 75th percentile is X when Z has a p-value of 0.25, so <u>X when Z = 0.675</u>.

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

$0.675 = \frac{X - 75}{5}$

$X - 75 = 0.675(5)$

$X = 78.4$

The 75th percentile for the math scores was of 78.4 inches.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B36%7D%7B4%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ \frac{36}{4}

iris [78.8K]

Step-by-step explanation:

$\sqrt{36/4} =$ $\sqrt{36} /\sqrt{4}$

$\sqrt{36}=6$

$\sqrt{4} =2$

6/2=3

Hope that helps :)

7 0

3 years ago