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3 years ago

Change the fraction 3/4 to a percent.

1 answer:

5 0

There is 100% in 1 since this is three parts of 1 we will have to divide 100 by 4 then multiply it by 3.
After you do all the math you will get 0.75.
You can convert this to a percent by just removing the point and the zero and adding a percent at the end.
The answer is going to be 75%

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Based on a 4-week work month and a 40-hour

Leni [432]

Answer:

$27.69

Step-by-step explanation:

Given

Monthly budget (See attachment)

$Month = 4\ weeks$

$Week = 40 \ hour$

Required

Determine the closest hourly wage

From the attachment, the monthly budget is:

$Budget = \$ 4430$

First, we calculate the closest weekly earnings.

Since a month = 4 weeks.

Then

$Weekly\ Earnings = \frac{\$4430}{4}$

$Weekly\ Earnings = \$1107.5$

Next, we calculate the hourly earnings:

Since 1 week = 40 hour

Then

$Hourly\ Earnings = \frac{\$1107.5}{40}$

$Hourly\ Earnings = \$27.6875$

<em>From the list of options, the closest is: $27.69</em>

4 0

2 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

Tony buys a pair of running shoes f 80$ he pays 9%sales tax how much does the pair of shoes cost in total?

Sindrei [870]

Answer:

$87.20

Step-by-step explanation:

5 0

3 years ago

Charles is 19 years old. his grandfather is 67 years older than he. what was thee total age 9 years ago?

almond37 [142]

Today: x=19
y=67+19=86

9 years ago: x=19-9=10
y=86-9=77

The total age 9 years ago: x+y=10+77=87

3 0

3 years ago

Read 2 more answers

Can i get help please i don't know this

scoundrel [369]

I think it means you have to find the area of both shapes and then compare them to see if Liam is correct if he is not correct say why. Hope this helps :)

3 0

3 years ago