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mrs_skeptik [129]
3 years ago
11

What times 16 equals 2048?

Mathematics
2 answers:
Zina [86]3 years ago
5 0

Answer:

128

hope this helps :)

Angelina_Jolie [31]3 years ago
4 0

Answer:128

Step-by-step explanation:

Divide 2048 by 16 yo get your answer

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drek231 [11]

Answer:its 10 :)

Step-by-step explanation:

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3 years ago
The hypotenuse of an isosceles triangles measures 10 inches long. What is the length of one leg of the triangle?
Ainat [17]

The answer is 10/√2 which would be A.


4 0
3 years ago
Read 2 more answers
One year, the population of a city was 397,000. Several years later it was 412,880.
artcher [175]

Answer:

4 percent increase in population

Step-by-step explanation:

percent increase = ((412,880 - 397,000)/ 397,000 ) × 100

7 0
3 years ago
An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were
vladimir1956 [14]

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.

3 0
3 years ago
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garri49 [273]
What am I suppose to be answering?
4 0
3 years ago
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