I need help so can someone help me please

Find the area of each sector. Photo attached

Flura [38]

Answer:

Area of sector: 48π in^2; Option D

Step-by-step explanation:

~ We can see that the bolded portion of the circle must be the the sector that we need to compute the area of... ~

Let us plan out this problem. If we were to find the area of the circle provided a radius of 12 inches, we could create a proportionality between the degree of the sector, and the degree of the circle ( 360 ) and the area of the sector and the area of the circle.

1. To calculate the area of this circle, let us apply the formula πr^2, such that r is 12 inches ( and let us keep π in terms of π ) ⇒ π * ( 12 )^2 ⇒ 144 * π ⇒ Area of Circle: 144π

2. Let us see this as a proportionality:

Degree of sector/ Degree of circle = Area of sector/ Area of circle

3. Now let us say x ⇒ Area of the sector. Substitute values into this formula, and solve for x through algebra:

120 / 360 = x / 144π,

120 * 144π = 360 * x,

17,280π = 360 * x,

x = 17,280π / 360,

Area of sector: 48π in^2

3 0

3 years ago

Find the area for the following figure.

IgorC [24]

Answer:

Step-by-step explanation: 3.5 times 1.7 =5.95

5 0

2 years ago

The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The

Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(c) The necessary assumption is that the population is normally distributed.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let X = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

(a)

The general form of a (1 - α)% confidence interval is:

$CI=SS\pm CV\times SD$

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a z-confidence interval.

The critical value of z for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the confidence interval as follows:

$CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)$

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

$37\pm 0.1=SS\pm CV\times SD$

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

$P(-1$

The percentage of z-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the z-interval or t-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a t-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For n = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - α)% t-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$