Help me find perimeter

Mathematics

1 answer:

aev [14]4 years ago

5 0

The perimeter is made up of the quarter arc and 2 radius.

Find the length of the arc:
Length of the arc = 1/4( 2πr) = 1/4 (2 x 22/7 x 28) = 44cm

Find the perimeter
Perimeter = 44 + 28 + 28 = 100cm

Answer; 100 cm

You might be interested in

Linda is making a line plot of the data below. 11.25, 12.5, 11.25, 14.125, 10.5, 11.25, 12 What are the least and greatest value

Morgarella [4.7K]

Least: 10.5

Greatest: 14.125

Hope that helps!

3 0

3 years ago

Y is a differentiable function of x. Choose the alternative that is the derivative dy / dx.

murzikaleks [220]

Differentiating both sides of

$x^3-y^3=1$

with respect to x yields (using the chain rule)

$3x^2 - 3y^2 \dfrac{\mathrm dy}{\mathrm dx} = 0$

Solve for dy/dx :

$3x^2 - 3y^2 \dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ 3y^2\dfrac{\mathrm dy}{\mathrm dx} = 3x^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{3x^2}{3y^2} = \dfrac{x^2}{y^2}$

The answer is then D.

3 0

3 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

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".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

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 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by: