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slava [35]
3 years ago
12

Please help solve the problem below.

Mathematics
1 answer:
ivann1987 [24]3 years ago
3 0

Answer:

1. x = 2

2. x = 5

3. x = 8

4. x = 28

5. x = 13

6. x = 11

7. x = 0

8. x = 72

9. x = 6

10. x = 38

Step-by-step explanation:

hope that helps!

1. subtract 6 on both sides

2. add 3 on both sides

3. divide both sides by 2

4. multiply both sides by 2

5. add 8 on both sides

6. subtract 9 on both sides

7. add 5 on both sides

8. multiply both sides by 6

9. divide both sides by 4

10. add 18 on both sides

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y = 3/4 x + ( - 2)
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A spherical ball with a volume of 972π in.3 is packaged in a box that is in the shape of a cube. The edge length of the box is e
aivan3 [116]
To find the volume of the box you will use the volume of the sphere formula to find the radius of the circular region of the sphere. You wil then double the radius to find the diameter.

V = 4/3 x pi x r^3
972 x pi= 4/3 x pi x r^3
729 = r^3
The cubic root of 729 is 9 because 9 x 9 x 9 = 729.

The radius is 9 inches, and the diameter is 18 inches.

The volume of the box is 18 in x 18 in x 18 in or 5832 in.³.
4 0
3 years ago
9. GIVEN: y(x + 1) = 51; y = 3 PROVE: x = 16​
ivann1987 [24]

Answer:

see explanation

Step-by-step explanation:

y(x + 1) = 51 ← substitute y = 3 into the equation

3(x + 1) = 51 ( divide both sides by 3 )

x + 1 = 17 ( subtract 1 from both sides )

x = 16

7 0
2 years ago
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manufacturing company produces digital cameras and claim that their products maybe 3% defective. A video company, when purchasin
alexdok [17]

Answer:

P(X>17) = 0.979

Step-by-step explanation:

Probability that a camera is defective, p = 3% = 3/100 = 0.03

20 cameras were randomly selected.i.e sample size, n = 20

Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97

Probability that more than 17 cameras are working P ( X > 17)

This is a binomial distribution P(X = r) nCr q^{r} p^{n-r}

nCr = \frac{n!}{(n-r)!r!}

P(X>17) = P(X=18) + P(X=19) + P(X=20)

P(X=18) = 20C18 * 0.97^{18} * 0.03^{20-18}

P(X=18) = 20C18 * 0.97^{18} * 0.03^{2}

P(X=18) = 0.0988

P(X=19) = 20C19 * 0.97^{19} * 0.03^{20-19}

P(X=19) = 20C19 * 0.97^{19} * 0.03^{1}

P(X=19) = 0.3364

P(X=20) = 20C20 * 0.97^{20} * 0.03^{20-20}

P(X=20) = 20C20 * 0.97^{20} * 0.03^{0}

P(X=20) = 0.5438

P(X>17) = 0.0988 + 0.3364 + 0.5438

P(X>17) = 0.979

The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch

6 0
3 years ago
ANSWER IF YOU KNOW THE ANSWER!!! I WILL GIVE BRAINLIEST BTW!!!!!!!!!!!!!
AURORKA [14]

im pretty sure its c

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