All categories

4 years ago

What is the measure of ABC and please show me how to do it

Mathematics

1 answer:

oksano4ka [1.4K]4 years ago

8 0

Arc length from A to C (red line) is the given inside angle times 2, so AC = 50*2 = 100 degrees.

A full circle is 360 degrees, so arc ABC (blue line) would be 360 - 100 = 260 degrees.

The answer is D.

You might be interested in

Shannon scores 36 points on 6 questions on her test what is the unit rate for one question?

VLD [36.1K]

Answer:

Uh I think 6

Step-by-step explanation:

7 0

3 years ago

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - α) % confidence interval for population proportion is:

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

The margin of error in this interval is:

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

The formula to compute the sample size is:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

(1)

Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

8 0

3 years ago

Use the definition of continuity and the properties of limits to show that the function

Zepler [3.9K]

Answer:

Applied the definition and the limit.

They had the same result, so the function is continuous.

Step-by-step explanation:

At function f(x) is continuous at x = a if:

$\lim_{x \to a} f(x) = f(a)$

In this question:

$f(x) = x^{2} + 5(x-2)^{7}$

At x = 3.

$\lim_{x \to 3} x^{2} + 5(x-2)^{7} = 3^{2} + 5(3-2)^{7} = 14$

$f(3) = 3^{2} + 5(3-2)^{7} = 14$

Since $\lim_{x \to 3} f(x) = f(3)$ , f(x) is continuous at x = 3.

7 0

3 years ago

Twenty five employees decide to chip in to buy a small refrigerator for the office. If the cost of the refrigerator is 275 how m

ira [324]

Answer:

Each employee will need to contribute $55.

Step-by-step explanation:

275 / 5 = 55

5 0

3 years ago

Read 2 more answers

Two different right cones are being considered by a design team to hold one liter (1000

ollegr [7]

Answer:

The surface area of Design A is smaller than the surface area of Design B.

The area of Design A is 94.65% of the Design B.

Step-by-step explanation:

The area of a right cone is given by the sum of the circle area of the base and the lateral area:

$A = \pi r^{2} + \pi rL$ (1)

Where:

r: is the radius

L: is the slant height

The slant height is related to the height and to the radius by Pitagoras:

$L^{2} = H^{2} + r^{2}$

$L = \sqrt{H^{2} + r^{2}}$ (2)

By entering equation (2) into (1) we have:

$A = \pi r^{2} + \pi r(\sqrt{H^{2} + r^{2}})$

Now, let's find the area of the two cases.

Design A: height that is double the diameter of the base, H= 2D = 4r

$A_{1} = \pi r^{2} + \pi r(\sqrt{(4r)^{2} + r^{2}}) = \pi r^{2}(1+ \sqrt{17})$

The volume of the cone is:

$V = \frac{1}{3}\pi r^{2}H$

We can find "r":

$V = \frac{1}{3}\pi r^{2}(4r) = \frac{4}{3}\pi r^{3}$

$r = \sqrt[3]{\frac{3V}{4\pi}} = \sqrt[3]{\frac{3*1000}{4\pi}} = 6.20 cm$

The area is:

$A_{1} = \pi (6.20)^{2}(1+ \sqrt{17}) = 618.7 cm^{2}$

Design B: height that is triple the diameter of the base, H = 3D = 6r

The radius is:

$r = \sqrt[3]{\frac{3V}{6\pi}} = \sqrt[3]{\frac{3*1000}{6\pi}} = 5.42 cm$

The area is:

$A_{2} = \pi r^{2} + \pi r(\sqrt{(6r)^{2} + r^{2}}) = \pi r^{2}(1 + \sqrt{37}) = \pi (5.42)^{2}(1 + \sqrt{37}) = 653.7 cm^{2}$

Hence, the surface area of Design A is smaller than the surface area of Design B.

The percent of the surface area of Design A is less than Design B by:

$\% A = \frac{618.7 cm^{2}}{653.7 cm^{2}}\times 100 = 94.65 \%$

Therefore, the area of Design A is 94.65% of the Design B.

I hope it helps you!

3 0

3 years ago