All categories

1 year ago

Your hourly salary increased from $13 to $20. What is the percent increase in your salary?

Mathematics

1 answer:

iVinArrow [24]1 year ago

8 0

The percent increase in your salary is 53.8%

<h3>What is the percent increase in your salary?</h3>

The given parameters are:

Initial = $13

Final = $20

The percent increase in your salary is calculated as:

Percent Increase = (Final - Initial)/Initial * 100%

So, we have:

Percent Increase = (20- 13)/13* 100%

Evaluate

Percent Increase = 53.8%

Hence, the percent increase in your salary is 53.8%

You might be interested in

A person places $51200 in an investment account earning an annual rate of 5.4%, compounded continuously. Using the formula V = P

serg [7]

Answer:

<em>7,078,912 cents</em>

Step-by-step explanation:

Given the formula for calculating the value of the account in t years as;

V = Pe^rt

P is the principal initially invested

e is the base of a natural logarithm,

r is the rate of interest

t is the time

Given

P = $51200

r = 5.4% = 0.054

t = 6years

Substitute

V = 51200e^(0.054)(6)

V = 51200e^(0.324)

V = 51200(1.3826)

V = $70,789.12

V = 7,078,912 cents

<em>hence the amount in the account after 6 years to the nearest cent is 7,078,912 cents</em>

7 0

3 years ago

Read 2 more answers

Tanya's parents gave her $500 for her high school graduation. She put the money into a savings account that earned 7.5% annual i

likoan [24]

Answer:

$567.5

Step-by-step explanation:

7.5 × 9 is 67.5 and Plus $500 would be $567.5

3 0

3 years ago

If cos 0=sin b then which of the following must be true?

miv72 [106K]

Rgtef erggreedg wrrevgerv erevt

7 0

4 years ago

Triangle ABC has the following vertices:

Zielflug [23.3K]

Answer:

I don't know ok bye bye see you

5 0

3 years ago

Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mea

QveST [7]

A.
$\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)$

Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.

$\mathbb P(Z>-2)=\mathbb P(|Z|2)=0.95+0.025=0.975$

b.
$\mathbb P(1100$

About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.

c. You're looking for $k$ such that

$\mathbb P(X>k)=0.10$

Since

$\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10$

occurs for $k^*\approx1.2816$ , it follows that

$\dfrac{k-1200}{100}=1.2816\implies k\approx1328$

So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.

3 0

4 years ago