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sergeinik [125]
3 years ago
14

Evaluate the following expression using the values given.

Mathematics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

-19

Step-by-step explanation:

2×2-2z4+y2-x2+z4

now we put the value,

4-2(2to the power 4)+(3 to the power 2)-(-4 to the power 2)+(2 to the power 4)

4-2×16+9-16+16

4-32+9

4+9-32

13-32

-19

please check the answer again i am sorry if it's wrong

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Order them from least, to greatest value.
ladessa [460]
-4
0.9
sqrt 85 = 9.219
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least to greatest : -4, 0.9, sqrt 85, 11, 4(pi)

lol...there already in order...just like u posted them
6 0
4 years ago
Solve for x . Round to the nearest tenth, if necessary.
Jet001 [13]

Answer:

x = 24.1

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

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90 tan 15 =x

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3 years ago
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3 years ago
A sample of 900 computer chips revealed that 61% of the chips fail in the first 1000 hours of their use. The company's promotion
Vlad [161]

Answer:

The p-value of the test is 0.0301 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

The company's promotional literature claimed that under 64% fail in the first 1000 hours of their use.

At the null hypothesis, we test if the proportion is of at least 64%, that is:

H_0: p \geq 0.64

At the alternative hypothesis, we test if the proportion is of less than 64%, that is:

H_1: p < 0.64

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

64% is tested at the null hypothesis:

This means that \mu = 0.64, \sigma = \sqrt{0.64*0.36}

A sample of 900 computer chips revealed that 61% of the chips fail in the first 1000 hours of their use.

This means that n = 900, X = 0.61

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.61 - 0.64}{\frac{\sqrt{0.64*0.36}}{\sqrt{900}}}

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P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.61, which is the p-value of z = -1.88.

Looking at the z-table, z = -1.88 has a p-value of 0.0301.

The p-value of the test is 0.0301 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the company's claim.

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3 years ago
Anyone have answers?
Anarel [89]

Answer:

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7 0
4 years ago
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