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raketka [301]
3 years ago
14

Help MEEEEEE PLEASEEEEEEE

Mathematics
2 answers:
Anton [14]3 years ago
4 0

Answer:

4 1/3 hours

Step-by-step explanation:

Since he will have completed 7 hours AFTER 2 2/3 more hours, he has completed 7- 2 2/3 = 4 1/3 hours so far

nasty-shy [4]3 years ago
4 0
The answer would be 4 1/3 hours
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Semenov [28]
Your answers are a and d hope this helps
4 0
4 years ago
I need help on 12, I don't know how to make that an expression! Thanks! Please explain!
exis [7]
A rug maker is using a pattern that is a rectangle with a length of 96 inches and a width of 60 inches. The rug maker wants to increase each dimension by a different amount. Let l and w be the increases in the length and width. Write and simplify an expression for the perimeter of the new pattern.

p=2(96+l)+2(60+w)
This is the equation. p is for perimeter. (96+l) represents the original length plus the change in length. The 2 before (96+l) represents that there is one length on each side of the rectangle.
Same for the width. (60+w) represents the original width plus the change in width. The 2 before (60+w) represents that there is one width on each side of the rectangle.
The simplified equation is p=(192+2l)+(120+2w) (this is your answer)

I hope this helps!

3 0
3 years ago
43.586 rounded to the nearest hundredth
garik1379 [7]
So you round to the second decimal place, giving you 43.59
3 0
3 years ago
Read 2 more answers
0.249 in expanded form
nikklg [1K]
2.49×10^-1 is the answer
8 0
3 years ago
Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce
34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(boxes weigh more than 22.9 ounces)

P(x > 22.9)

P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)

= 1 - P(z \leq -2)

Calculation the value from standard normal z table, we have,  

P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 23.7) =0.159= 15.9\%

15.9% of of the boxes weigh less than 23.7 ounces.

7 0
3 years ago
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