If 25% of a number p is 19, what is 15 of p

X2 2 ab ab to b The diagian K n shows rectangle. The shade Find x.

Norma-Jean [14]

Answer:

x = 28 cm

Step-by-step explanation:

Given:

Area of link shaded regions = 84 cm²

Required:

The value of x (diameter of the semicircle/length of the rectangle)

Solution:

Diameter of the semicircle = 2r = x

Length of rectangle (L) = 2r = x

Radius of semicircle (r) = ½x

Width of rectangle (W) = radius of semicircle = ½x

Use 3.14 as π

Area of the link shaded regions = area of rectangle - area of semicircle

Thus:

Area of the link shaded regions = (L*W) - (½*πr²)

Plug in the values

84 = (x*½x) - (½*3.14*(½x)²)

84 = x²/2 - (1.57*x²/4)

84 = x²(½ - 1.57/4)

84 = x²(0.5 - 0.3925)

84 = x²(0.1075)

Divide both sides by 0.1075

84/0.1075 = x²

781.4 = x²

√781.4 = x

27.9535329 = x

x = 28 cm

5 0

3 years ago

My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: