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Mama L [17]
3 years ago
9

NO LINKS OR ELSE YOU'LL BE REPORTED!Only answer if you're very good at Math.No guessing please.

Mathematics
2 answers:
Yuki888 [10]3 years ago
5 0
C due to the fact that it is C and C is the correct answer
puteri [66]3 years ago
5 0

Answer:

16 3/2

Step-by-step explanation:

So lets go thru each one.

16 3/2 is just 17.5

A square root cancels out a square.

A cubed root cancels out a cube.

So, in the next answer:

16 squared, to the square root is just 16, since the squaring and root cancels out.

64 sqaured to the cube root must be 16. Lets break it all down:

2^6=64.  We need to square this, so we can rewrite it as:

64^2 = (2^6)^2

Normally we add/subtract exponents, but since the 6 exponent is in parethesese, it is multiplied by the 2 exponent. This makes it 2^12. So we can rewrite \sqrt[3]{64^2} as:

\sqrt[3]{2^1^2}

Now, when we divide a exponent by a exponent, we subtract.

However, when we root a exponent, it is division.

So it is the 2 \frac{^1^2}{^3}

=

2^4

Or

16

Finally we have the cubed root of 8^4.

using the same steps as above, 8^4 = (2^3)^4. This equals 2^1^2

Then again, we cube root it.

This is division of the 12 exponent by the 3 root, so we get:

2^4

=

16

So in the end:

A - 17.5

B - 16

C - 16

D - 16

So a is the largest number

Hope this helps!

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Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

<u>Step:-(1)</u>

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

<u>Step :-2</u>

<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

<u>Alternative hypothesis :H₁: </u>there is  significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

                     Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} }  } }

Given  n₁=n₂=60.

                    Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60}  } }

On calculation, we get

                   Z =   \frac{4}{\sqrt{0.6833} }

                   z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

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