Ask question

Ask question

All categories

3 years ago

13

The top and bottom ends of a windshield wiper blade are R = 24 in. and r = 14 in., respectively, from the pivot point. While in

operation, the wiper sweeps through 135°. Find the area swept by the blade. (Round your answer to the nearest whole number.)

1 answer:

Orlov [11]3 years ago

3 0

Answer:

Area swept by the blade = 448 $in^{2}$

Step-by-step explanation:

The arc the wiper wipes is for 135 degrees angle.

So, find area of sector with radius 24 inches. And the find area of arc with r=14 inches.

Then subtract the area of sector with 14 inches from area of sector with radius as 24 inches.

So, area of sector with r=24 in = $\frac{135}{360} *\pi *24^{2}$

Simplify it,

=216 $\pi$

Now, let's find area of sector with radius 14 inches

Area of sector with r=14 in = $\frac{135}{360} *\pi *14^{2}$

Simplify it

=73.5 $\pi$

So, area swept by the blade = 216 $\pi$ -73.5 $\pi$

Simplify it and use pi as 3.14.....

Area of swept =678.584 - 230.907

=447.6769

Round to nearest whole number

So, area swept by the blade = 448 $in^{2}$

You might be interested in

Slove for x 7x-22=4x+5

Vadim26 [7]

Answer: x = 9

Step-by-step explanation:

5 0

3 years ago

If there are 25 students in a class - 11 are guys and 14 are girls what is the probability that one of the students on the class

levacccp [35]

Answer:

0.44

Step-by-step explanation:

11/25 = 0.44 = 44%

6 0

3 years ago

Read 2 more answers

Find the exponential equation for the given information. Pls help

nikitadnepr [17]

Answer:

y = 12 $(2.9)^{x}$

Step-by-step explanation:

The standard form of the exponential function is

y = a $(b)^{x}$

Find a and b by substituting ordered pairs from the table into the equation

Using (0, 12 ), then

12 = a $b^{0}$ ( $b^{0}$ = 1 ), so

a = 12, then

y = 12 $b^{x}$

Using (1, 34.8 ) , then

34.8 = 12 $b^{1}$ ( divide both sides by 12 )

b = 2.9

Thus exponential function is

y = 12 $(2.9)^{x}$

3 0

3 years ago

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm

Vlad1618 [11]

Answer:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar x =512$ represent the sample mean

$\sigma=100$ represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

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

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

$Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

$25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}$

And we can find the critical value $z_{\alpha/2}$ like this:

$z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576$

And we know that on the right tail of the z score =2.576 we have $\alpha/2$ of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998= $\alpha/2$ , so then $\alpha=0.00498*2=0.009995 \approx 0.01$ , and then we can find the confidence like this:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%

AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

7 0

3 years ago