A woman earning $20 an hour gets a 15% raise. With her raise, how much will she earn after working 40 hours?

Mathematics

1 answer:

erastovalidia [21]3 years ago

4 0

She will earn $920 after working 40 hours.

Step-by-step explanation:

Given,

Amount per hour = $20

Raise = 15%

Raise amount = 15% of Amount per hour

$Raise\ amount = \frac{15}{100}*20\\Raise\ amount= \frac{300}{100}\\Raise\ amount= \$3$

New amount per hour = Amount per hour + raise amount

New amount per hour = 20+3= $23

Total earning after working 40 hours = Amount per hour * 40

$Total\ earning=23*40\\Total\ earning= \$920$

She will earn $920 after working 40 hours.

Keywords: Raise, percentage

Learn more about percentages at:

#LearnwithBrainly

You might be interested in

(a)<br> x2 + 2x - 12 when x = 5

tiny-mole [99]

Answer:

Step-by-step explanation:

(5*2)+(2*5)-12=

10+10-12=

20-12=

8 0

3 years ago

At a large university, the mean amount spent by students for cell phone service is $58.90 per month with a standard deviation of

zhenek [66]

Answer:

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

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 are 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 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 $s = \frac{\sigma}{\sqrt{n}}$ .