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aalyn [17]
3 years ago
12

Kevin is paid $8.80 per hour. He worked 7 hours. He gave his mother 1/4 of his earnings. How much did he have left over?

Mathematics
2 answers:
Colt1911 [192]3 years ago
5 0

Answer:

$46.20

Step-by-step explanation:

His earnings: 8.8*7 = 61.6

He gave away 1/4, so he has 3/4 left

3/4*61.6=46.2

dusya [7]3 years ago
4 0

Answer:

$46.2

Step-by-step explanation:

If kevin is paid $8.80 per hour and he worked 7 hours, we need to multiply the hourly pay by the number of hours worked:

$8.80(7) = $61.6

In total I earn the amount of $61.6.

He gave 1/4 of this earnings ti his mother.

1/4 of $61.6 is:

\frac{61.6}{4} =15.4

so he gave $15.4 to his mother.

what he has left is the subtraction of what he earned and what he gave his mother:

$61.6 - $15.4  = $46.2

He has $46.2 left

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Ruby uses 1/4 yard of ribbon to make 2 bows.Which expression shows the length of ribbon in each bow?
allochka39001 [22]
What you can do in this case is a rule of three to determine the length of each bow.
 We have then:
 1/4 ---> 2
 x ------> 1
 Clearing x we have:
 x = (1/2) * (1/4)
 x = 1/8
 Answer:
 the length of ribbon in each bow is
 x = 1/8
 Equivalently:
 x = (1/4) / 2
 Option 3
3 0
3 years ago
NEED HELP FAST! ALSO 50 POINTS AND BRAINLIEST!! <br><br> What is the value of 5 2/3 (−3.6)?
ratelena [41]

Answer:

-20⅖

Step-by-step explanation:

5⅔(-3.6)

(17/3)(-3.6)

17(-3.6/3)

17(-1.2)

-20.4

-20 ⅖

8 0
3 years ago
In a class of 50 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose
Tems11 [23]

Answer:  3

Step-by-step explanation:

Let E be the event of that student pierces ear and N be the event of that student pierces nose.

Given: n(E\cup N=50)

n(E)=46\\\\n(N)=7

For any two event A and B, we have

n(A\cup B)=n(A)+n(B)-n(A\cap B)

Similarly , n(E\cup N)=n(E)+n(N)-n(E\cap N)

50=46+7-n(E\cap N)\\\\\Rightarrow\ n(E\cap N)=53-50=3

Hence, 3 students have piercings both on their ears and their noses.

7 0
3 years ago
One sample has a mean of and a second sample has a mean of . The two samples are combined into a single set of scores. What is t
Murrr4er [49]

Answer:

a) For this case we can use the definition of weighted average given by:

M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

And if we replace the values given we have:

M = \frac{8*4 + 16*4}{4+4}= 12

b) M = \frac{8*3 + 16*5}{3+5}= 13

c) M = \frac{8*5 + 16*3}{5+3}= 11

Step-by-step explanation:

Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.

a) What is the mean for the combined set if both of the original samples have n=4 scores "

For this case we can use the definition of weighted average given by:

M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

And if we replace the values given we have:

M = \frac{8*4 + 16*4}{4+4}= 12

b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5

Using the definition we have:

M = \frac{8*3 + 16*5}{3+5}= 13

c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3

Using the definition we have:

M = \frac{8*5 + 16*3}{5+3}= 11

7 0
3 years ago
Question 4 of 15
melisa1 [442]
“C” Os the correct answer
5 0
3 years ago
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