All categories

3 years ago

If the population standard deviation is 10 , how large a sample is necessary to have a standard error that is

1 answer:

5 0

The standard error given a population standard deviation of $\sigma$ is $\dfrac\sigma{\sqrt n}$ . So if the desired standard error is $s$ , then

$s=\dfrac\sigma{\sqrt n}\implies \sqrt n=\dfrac\sigma s\imlies n=\dfrac{\sigma^2}{s^2}$

You might be interested in

In algebra 1 how do I solve for y in this problem? -3x=y+2

Alenkinab [10]

Y=−3x−2
-3x=y+2 (keep, change, flip)
so....
-3x-2 :) Hope I helped!

5 0

3 years ago

An earthquake measuring 6.4 on the Richter scale struck Japan in July 2007, causing extensive damage. Earlier that year, a minor

a_sh-v [17]

Answer:

Step-by-step explanation

The earthquake measures 6.4 on the Richter scale which struck Japan in Jullu 2007 and caused and extensive damage. Earlier that year, a minor earthquake measuring 3.1 in the Richter scale has stroked in parts of Pennsylvania.

Fomular:

The magnitude of an earthquake is M log(I/S)

where I donates the intensity of the earthquake and S be the intensity of the standard earthquake.

Calculation:

Consider that M1 be the magnitude Japanese earthquake and M2 be the magnitude of the Pennsylvania earthquake and L1 be the intensity of the Japanese earthquake and L2 the intensity of the Pennsylvania earthquake.

Here the magnitude of the Japanese earthquake is M1 = 6.14 and the magnitude of the Pennsylvania is M2 = 3.1

By the use of magnitude of the earthquake fomular M = log I1/S, the intensity of the Japanese earthquake is calculated as follows .

M1 = log I1/S

I1/s = 10

3 0

3 years ago

I need homework help! Its proportions, but the thing is, I'm terrible at them. All I know is that is/of=x/100.

masha68 [24]

19) 15/a = 3/2

Start by cross multiplying....

3a = 30

Divide both sides 3

a = 10

21) 2/7 = 4/d

2d = 28

   d= 14

23) 8/p = 3/10

80 = 3p
p = 26.66

25) 2 / -5 = 6/t
  2t = -30

  t = -15

7 0

3 years ago

change the precent to a fraction 120% (simplify your answer) show work with this and steps on the side

storchak [24]

Here are my steps 11/5= 2 1/5

4 0

3 years ago

1. A circle with center (-5, 2) is tangent to the y-axis. What is the radius of the circle? What is the equation of the circle?

Bess [88]

Answer:

Step-by-step explanation:

1) The center lies on the vertical line x = -5 and the the circle is tangent to (touches in one place only) the y-axis. Thus, the radius is 5.

2) Starting with (x - h)^2 + (y - k)^2 = r^2 and comparing this to the given

(x - 4)^2 + (y + 3)^2 = 6^2

we see that h = 4, k = -3 and r = 6. The center is at (4, -3) and the radius is 6.

3) Notice that A and B have the same x-coordinate, x = 15. The center of the circle is thus (15, -2), where that -2 is the halfway point between the two given points in the vertical direction. Arbitrarily choose A(15, 4) as one point on the circle. Then the equation of this circle is

(x - 4)^2 + (y + 3)^2 = r^2 = 6^2, where the 6 is one half of the vertical distance between A(15, 4) and B(15, -8) (which is 12).

4 0

2 years ago