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4 years ago

5

A car's rear windshield wiper rotates 140°. The total length of the wiper mechanism is 24 inches and the length of the wiper bla

de is 16 inches. Find the area wiped by the wiper blade. (Round your answer to one decimal place.)

1 answer:

Verizon [17]4 years ago

6 0

Answer:

<h2>312.7 in^2</h2>

Step-by-step explanation:

This problem is on the mensuration of flat shapes, sectors and arc.

we know that the area of a sector is given as

A=(∅/360)*πr^2

Given that the length of the wiper R=24 in

and the length of the blade r= 16in

To find the area of the wiped surface we only require the length of the blade as the radius

The area of the wiped surface is given as= (140/360)*3.142*16^2

A= 0.3888*3.142*256

A=312.7 in^2

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4k + 15 greater than -2k + 3

ioda

Answer:

k>-2

Step-by-step explanation:

4k+15>-2k+3

4k-(-2k)+15>3

4k+2k+15>3

6k+15>3

6k>3-15

6k>-12

k>-12/6

k>-2

8 0

3 years ago

A helicopter takes off from the ground 800 feet from the base of a building. It flies in a straight line directly to the top of

Stolb23 [73]

Answer:

11.3°

Step-by-step explanation:

Refer the attached figure .

Height of building i.e. AB = 160 feet.

A helicopter takes off from the ground 800 feet from the base of a building i.e. BC = 800 ft.

We are required to find what angle did the helicopter take off from i.e. ∠ACB

We will use trigonometric ratio .

$tan\theta = \frac{Perpendicular}{Base}$

$tan\theta = \frac{AB}{BC}$

$tan\theta = \frac{160}{800}$

$tan\theta = \frac{1}{5}$

$tan\theta = 0.2$

$\theta =tan^{-1} 0.2$

$\theta =11.3^{\circ}$

Thus the angle at which the helicopter take off from is 11.3°

3 0

4 years ago

How do u find best fit line

Sliva [168]

It is b because you can reason and envision a line that goes through the points.

7 0

4 years ago

Find the next term in the following number pattern: 1, 16, 81, 256, 625, ____

Helen [10]

1296

1^4 = 1
2^4 = 16
3^4 = 81
4^4 = 256
5^4 = 625
6^4 = 1296

3 0

2 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

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 can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago