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3 years ago

What is 2 to the second power times 2 to the third power

Mathematics

2 answers:

aleksandrvk [35]3 years ago

6 0

2*2= 4
2*2*2 = 8
4*8=32
So, your answer would be 32. Hope this helps! Please select as brainliest answer!

Hoochie [10]3 years ago

3 0

2^2 x 2^3 = 2^5
When multiplying with exponents that have the same base, you simply add the exponents and leave the base as it is.

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Name six pairs of supplementary angles

Bas_tet [7]

Supplementary angles are those angles whose sum of the measures of angles is 180 degrees.

Observe the figure clearly.

1. $\angle AGF , \angle AGC$ are supplementary angles.

2. $\angle FGD, \angle DGC$ are supplementary angles.

3. $\angle FGE, \angle EGD, \angle DGC$ are supplementary angles.

4. $\angle AGE, \angle AGB$ are supplementary angles.

5. $\angle BGC, \angle CGD, \angle DGE$ are supplementary angles.

6. $\angle CGE, \angle EGF$ are supplementary angles.

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3 years ago

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3 years ago

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Alenkinab [10]

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3 years ago

How do I solve 15+5-5/12. please give me the steps

gregori [183]

1) You must do division first. So divide 5/12 and get .41666666...
2) Then you add 15+5 and get 20.
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7 0

3 years ago

A sample of 5 buttons is randomly selected and the following diameters are measured in inches. Give a point estimate for the pop

Helen [10]

Answer:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

$\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072$

And for the sample variance we have:

$s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003$

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

$E(s^2) = \sigma^2$

Step-by-step explanation:

For this case we have the following data:

1.04,1.00,1.13,1.08,1.11

And in order to estimate the population variance we can use the sample variance formula:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

$\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072$

And for the sample variance we have:

$s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003$

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

$E(s^2) = \sigma^2$

3 0

3 years ago