Ask question

Ask question

All categories

Marina CMI [18]

3 years ago

8

Kyle works 40 hours at a rate of $10.75 per hour. He has total deductions of $87.25. What is his net pay? A. $430.00 B. $342.75

C. $362.25 D. $412.75

2 answers:

ivolga24 [154]3 years ago

6 0

B.342.75. 40 x 10.75 = 430 - 87.25= 342.75

AleksAgata [21]3 years ago

3 0

Multiply his hourly wage by the amount of hours he worked:

$40 \times 10.75 = 430$

Subtract the total deductions from the amount he earns to find the net pay:

$430 - 87.25 = 342.75$

Kyle's net pay is $342.75.

You might be interested in

Find the product using model or drawing

antiseptic1488 [7]

Fraction multiplication involves multiplying the numerators and the denominators to get the numerator and the denominator of the product respectively

Please find the attached diagrams

1. $\dfrac{2}{15}$ , 2. $\dfrac{3}{8}$ , 3. $\dfrac{1}{12}$ , 4. $\dfrac{7}{21}$ , 5. $\dfrac{3}{10}$

6. 51, 7. 1, 8. 22, 9. $\dfrac{36}{55}$ , 10. $5\dfrac{1}{3}$

Reasons:

The process of multiplication of factors using a model includes;

Creating a table that have the number of rows and columns equal to the values in the denominator of the two fractions respectively

Shading the number of columns given by the numerator of the fraction that has a denominator equal to the number of columns in the table

Shading the number of rows given by the numerator of the fraction that has a denominator equal to the number of rows in the table

Ensuring that the common area are shaded to a different color

Simplifying the result

1. $\dfrac{2}{3} \times \dfrac{1}{5} = \dfrac{2}{15}$

2. $\dfrac{1}{2} \times \dfrac{3}{4} = \dfrac{3}{8}$

3. $\dfrac{1}{4} \times \dfrac{2}{6} = \dfrac{2}{24} = \dfrac{1}{12}$

4. $\dfrac{1}{7} \times \dfrac{2}{3} = \dfrac{7}{21}$

5. $\dfrac{4}{8} \times \dfrac{3}{5} = \dfrac{12}{40} = \dfrac{3}{10}$

Second part

6. When calculating fractions, 'of' means to 'multiply by', therefore, we have;

$\dfrac{17}{25} \times 75= \dfrac{17}{25} \times 25 \times 3 = 17 \times 3 = 51$

7. $\dfrac{7}{6} \times \dfrac{24}{28} = \dfrac{7}{6} \times \dfrac{6 \times 4}{7 \times 4} = \dfrac{7}{6} \times \dfrac{6}{7 } = 1$

8. $12 \times 1\dfrac{5}{6} =6 \times 2 \times \dfrac{11}{6} =2 \times {11} = 22$

9. $\dfrac{4}{5} \times \dfrac{9}{11} = \dfrac{4 \times 9}{5 \times 11} = \dfrac{36}{55}$

10. $6\dfrac{2}{3} \times \dfrac{12}{15} =\dfrac{20}{3} \times \dfrac{12}{15} = \dfrac{4 \times 5}{3} \times \dfrac{4 \times 3}{5 \times 3} = \dfrac{4}{3} \times {4} = \dfrac{16}{3} = 5\dfrac{1}{3}$

Learn more here:

brainly.com/question/2752457

7 0

3 years ago

In October 1945, the Gallup organization asked 1,487 randomly sampled Americans, "Do you think we can develop a way to protect o

GalinKa [24]

Answer:

Yes. There is enough evidence to support the claim that the proportion of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945 is significantly greater than 0.5.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>In October 1945, the Gallup organization asked 1487 randomly sampled Americans, "Do you think we can develop a way to protect ourselves from atomic bombs in case other countries tried to use them against us?"</em><em> with 788 responding yes.</em><em> Did a majority of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945? Use the α=0.05 level of significance.</em>

This is a hypothesis test for a proportion.

The claim is that the proportion of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945 is significantly greater than 0.5.

Then, the null and alternative hypothesis are:

H_0: \pi=0.5\\\\H_a:\pi>0.5

The significance level is 0.05.

The sample has a size n=1487.

The sample proportion is p=0.53.

$p=X/n=788/1487=0.53$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{1487}}\\\\\\ \sigma_p=\sqrt{0.000168}=0.013$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.53-0.5-0.5/1487}{0.013}=\dfrac{0.03}{0.013}=2.288$

This test is a right-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z>2.288)=0.011$

As the P-value (0.011) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945 is significantly greater than 0.5.

8 0

3 years ago

Use substitution to find the solution of the system of equations. 5x – y = 17 6x + 2y = 14

NikAS [45]

2(5x - y = 17)
10x - 2y = 34
6x + 2y = 14
16 x = 48
x = 3

5(3) - y = 17
15 - y = 17
-y = 2
y = - 2

( 3 , -2)

5 0

4 years ago

Solve 13 + x> 11. Enter your answer as an inequality.

In-s [12.5K]

Answer:

x > -2

Step-by-step explanation:

Simply subtract 13 from both sides to find you answer!

4 0

4 years ago

Read 2 more answers

Calculate the HCF of 8a³-27, 4a²-9 and 4a²-12a+9

Lyrx [107]

$\: \: \: \: \:$

first add

(3a²-5a+5)+(a³-5a²+8)

=3a²-5a+5+a³-5a²+8

=a³-2a3a²-5a²+5+8

=a³-2a²+13

and now,

substract a³-2a²+13 from a²-4a²-9

=7a²-4a²-9-(a³+13)

=7a²-4a²-9-a³+2a²-13

= a³+7a²-4a²+2a²-9-13

=-a³+5a²-22

<h2><u>hope</u><u> it</u><u> helps</u></h2>

7 0

3 years ago