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3 years ago

Find the perimeter of a rectangle with a length of (attachment 1) and a width of(attachment 2) .

Mathematics

1 answer:

son4ous [18]3 years ago

5 0

$P_{\fbox{ }}=2w+2l\\\\w=x^2+2x+7;\ l=-2x+8\\\\P_{\fbox{ }}=2(x^2+2x+7)+2(-2x+8)=2x^2+4x+14-4x+16\\\\\boxed{P_{\fbox{ }}=2x^2+30}$

You might be interested in

Use the diagram. AB is a diameter, and AB is perpendicular to CD. The figure is not drawn to scale.

Westkost [7]

Answer:

The measure of arc BD is 147°

Step-by-step explanation:

we know that

If segment AB is perpendicular to segment CD

then

The measure of the inner angle CPA is a right angle (90 degrees)

Remember that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

m∠CPA=(1/2)[arc AC+arc BD]

substitute the given values

90°=(1/2)[33°+arc BD]

180°=33°+arc BD

arc BD=180°-33°=147°

7 0

3 years ago

What is the area of this quadrilateral

sammy [17]

Answer:

Step-by-step explanation:

First, let's work out the two triangles:

Area of triangle: bxh/2

2 x 8/2 = 8

Because there is two triangles multiply 8 by 2 = 16

To work out the area of the central rectangle:

Area of rectangle: l x w

3 x 8 = 24

Add 16+24 = 40 square feet

4 0

2 years ago

The slope of line m is 7/8. If n⊥m, what is the slope of line n?

Citrus2011 [14]

Hello!

Slopes of perpendicular lines are opposite reciprocals of each other, which the slope of line perpendicular to a line with a slope of 7/8 is -8/7.

Answer:
D. -8/7

Hope this helps!

8 0

3 years ago

There are 1,023 M&M's in a bag. Out of 1,023 M&M's, 132 M&M's are blue, 69 are green, 68 are brown, 63 are olive, 62

lesantik [10]

The probability that the person pulls 1 blue M&M given that one is only allowed to pull 5 random M&M's per day will be 0.6450.

<h3>How to calculate probability?</h3>

From the information given, there are 1,023 M&M's in a bag and out of 1,023 M&M's, 132 M&M's are blue.

Therefore, the probability of getting a blue M&M will be:

= 132/1023

= 0.1290

Therefore, since the person is given the chance to pull 5 random M&M's per day, the probability will be:

= 0.1290 × 5

= 0.6450

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

4 0

2 years ago

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

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".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that $t_{\alpha/2}=$

Now we have everything in order to replace into formula (1):

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

5 0

3 years ago