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Black_prince [1.1K]
4 years ago
15

4(960+2-125)+5 what is the solution to this mathematical expression?

Mathematics
2 answers:
seropon [69]4 years ago
5 0

Answer:

3353

Step-by-step explanation:

1. Simplify 960 + 2 to 962.

4 · (962 − 125) + 5

2. Simplify 962 − 125 to 837.

4 · 837 + 5

3. Simplify 4 · 837 to 3348.

3348 + 5

4. Simplify

3353

mezya [45]4 years ago
5 0

Answer: 3353

Step-by-step explanation:

4 × 960 = 3840

4 × 2 = 8

4 × -125 = -500

-500 + 8 + 3840 = 3348

3348 + 5 = 3353

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A cube has a volume 7^3 of cubic centimeters. What is its surface area?
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294 cm^2

Step-by-step explanation:

length of the cube is 7cm each side

S.A = 2(7 * 7) + 2( 7 * 7) + 2( 7 * 7)

= 294 cm^2

4 0
3 years ago
What is 15.4+5.4 its very hard for me to learn after all home school​
creativ13 [48]

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15.4+5.4=20.8

5 0
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Shape ABCD is an arrowhead.<br> Use the X<br> tool and mark point D<br> accurately on the graph.
ehidna [41]

Answer:did you know i like frogs

Step-by-step explanation: because i do

7 0
3 years ago
The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

           \sigma   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

7 0
3 years ago
HELPPPPP AGAIN PLZZZ!!!! :)
posledela
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ volume = length x width x height

volume = 3.5 x 1.5 x 2

volume = 10 1/2 cubic inches

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

6 0
3 years ago
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