All categories

4 years ago

Quan's last paycheck shows that he eamed $210.00 for 24 hours of work. What is his hourly rate of pay?

Mathematics

1 answer:

bija089 [108]4 years ago

4 0

Answer: 8.75$

Step-by-step explanation: to find the answer to this problem, you would divide his total amount of cash earned by the number of hours he worked to see how the cash is split between each hour. By doing this, you should get 8.75$ and hour.

You might be interested in

The speed limit on a rural road in Europe is 90 km / hr . Yvonne is driving 35 miles How long will it take her if she drives at

Daniel [21]

63 dnjdjdjdjdjdjdhdhhdhdhehusushshsbsbdddu

4 0

3 years ago

The price of gas, which started at $2.25 per gallon, increased at a rate of 4% per year. Write the function that models the situ

Rina8888 [55]

Answer:

$f ( x ) = 2 . 2 5 ( 1 + 0 . 4 ) ^ x$

Step-by-step explanation:

We are given that the price of gas started at $2.25 and increased at a rate of 4% per year.

We are to write a function which models this situation using numerical values of A and r into the expression.

A = $2.25

r = 4%

Assuming x to be the time in years, the function would be:

$f ( x ) = 2 . 2 5 ( 1 + 0 . 4 ) ^ x$

6 0

3 years ago

Read 2 more answers

Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance

lana66690 [7]

Answer:

Null Hypothesis, $H_0$ : p = 0.20

Alternate Hypothesis, $H_a$ : p > 0.20

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20% of users develop nausea.

Let p = population proportion of users who develop nausea

So, Null Hypothesis, $H_0$ : p = 0.20

Alternate Hypothesis, $H_a$ : p > 0.20

Here, null hypothesis states that 20% of users develop nausea.

And alternate hypothesis states that more than 20% of users develop nausea.

The test statistics that would be used here is One-sample z proportion test statistics.

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of users who develop nausea in a sample of 241 subjects = $\frac{54}{241}$

n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.

6 0

4 years ago

If x = 6 and y= 4, evaluate the following expression: 3x^2 + 3xy + y^2

Shkiper50 [21]

Answer:

plz refer the attachment

4 0

4 years ago

Read 2 more answers

2. Suppose there are 6 million people in New York City that walk an average of 2.5 miles per day. If we add up all the miles wal

Lunna [17]

Answer:

600 times

Step-by-step explanation:

<h3>Given</h3>

Number of people:

6 million = 6*10^6

Distance to walk:

2.5 miles per day per person

Distance around the Earth

2.5*10^4

<h3>Solution</h3>

Total distance walked:

6*10^6*2.5 = 15*10^6

The ratio of total distance to the distance around the Earth:

15*10^6 / 2.5*10^4 = 6*10^2 = 600 times

8 0

3 years ago

Read 2 more answers