The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. They would like the estimate to have a maximum error of 0.09 kWh. A previous study found that for an average family the variance is 5.76 kWh and the mean is 16.6 kWh per day. If they are using a 98% level of confidence, how large of a sample is required to estimate the mean usage of electricity

A sample of 3851 is required.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 2.327.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

Variance is 5.76 kWh

This means that

They would like the estimate to have a maximum error of 0.09 kWh. How large of a sample is required to estimate the mean usage of electricity?

This is n for which M = 0.09. So

Rounding up:

A sample of 3851 is required.

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this