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3 years ago

Use the given information to find the minimum sample size required to estimate an unknown population mean.

Mathematics

2 answers:

Harrizon [31]3 years ago

4 0

Answer:

c or b

Step-by-step explanation:

trapecia [35]3 years ago

4 0

Answer:

C) 94

Step-by-step explanation:

Margin of error = critical value × standard error

ME = CV × SE

Assuming n > 30, we can approximate CV with a normal distribution. At P = 95%, CV = 1.96.

SE = σ / √n

SE = 593 / √n

Therefore:

120 = 1.96 × 593 / √n

n = 94

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Help needed fast!!!

kicyunya [14]

Answer:

The answer is A

Step-by-step explanation:

It has to exceed not be same or equal to

6 0

3 years ago

Read 2 more answers

Please help!!As soon as possible!!

pychu [463]

Answer: Their equations have different y-intercept but the same slope

Step-by-step explanation:

Since, the slope of the line passes through the points (2002,p) and (2011,q) is,

$m_1 = \frac{q-p}{2011-2002} = \frac{q-p}{9}$

Similarly, the slope of the line asses through the points (2,p) and (11,q) is,

$m_2 = \frac{q-p}{11-2} = \frac{q-p}{9}$

Since, $m_1 = m_2$

Hence, both line have the same slope.

Now, the equation of the line one having slope $m_1$ and passes through the point (2002,p) is,

$y - p = \frac{q-p}{9}(x-2002)$

Put x = 0 in the above equation,

We get, $y = \frac{2000(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2000(p-q)}{9}+p)$

Also, the equation of second line having slope $m_2$ and passes through the point (2,p)

$y-p=\frac{q-p}{9}(x-2)$

Put x = 0 in the above equation,

We get, $y = \frac{2(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2(p-q)}{9}+p)$

Thus, both line have the different y-intercepts.

⇒ Third option is correct.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%

AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

7 0

3 years ago

To obtain the area of a sector, what fraction is multiplied by the area of a circle (A = πr2)?

romanna [79]

Let r be a radius of a given circle and α be an angle, that corresponds to a sector.

The circle area is $A=\pi r^2$ and denote the sector area as $A_1$ .
Then $\dfrac{A_1}{A}= \dfrac{\alpha}{2\pi}$ (the ratio between area is the same as the ratio between coresponding angles).

$A_1=\dfrac{\alpha}{2\pi} \cdot A=\dfrac{\alpha}{2\pi} \cdot \pi r^2= \dfrac{r^2\alpha}{2}$ .

6 0

3 years ago

If the measure of one angle in a pair of is n, the<br> The measure of the other angle is 180 – n.

olchik [2.2K]

This kinda doesn’t make sense but maybe 80, 80, 20

8 0

3 years ago