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kakasveta [241]
2 years ago
5

Marcie wants to enclose her yard with a fence. Her yard is in the shape of a rectangle attached to a triangle. The formula for t

he area of the enclosed space is A = lw + 0.5bh. Solve for b .
a. b = A divided by the quantity l times w plus 0.5 times h B . b = The quantity A minus l times w all divided by 0.5 times h
c. b = A – lw – 0.5h
d. b = A + lw + 0.5h No Idea :o
Mathematics
2 answers:
nydimaria [60]2 years ago
8 0

Let

l------> the length side of the rectangle

w----> the width side of a rectangle

b----> the base of the triangle

h----> the height of the triangle

we know that

the area of the yard is equal to

A=lw+0.5bh

Solve for b------> ( that means clear variable b)

Subtract lw both sides

A-lw=lw+0.5bh-lw

A-lw=0.5bh

Divide by 0.5h both sides

(A-lw)/(0.5h)=0.5bh/(0.5h)

b=(A-lw)/(0.5h)

therefore

<u>the answer is the option B</u>

The quantity A minus l times w all divided by 0.5 times h


DerKrebs [107]2 years ago
8 0
A=lw+0.5bh \ \ \ \ \ \ \ \ \ \ |-lw \\&#10;A-lw=0.5bh \ \ \ \ \ \ \ \ \ \ |\div 0.5h \\&#10;\boxed{b=\frac{A-lw}{0.5h}}

The answer is B.
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Answer:

11 km/hr.

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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:

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(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:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

(\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

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