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amm1812
3 years ago
9

Working on Summer Vacation. An Adweek/Harris (July 2011) poll found that 35% of U.S. adults do not work at all while on summer v

acation. In a random sample of 10 U.S. adults, let x represent the number who do not work during summer vacation
a. For this experiment, define the event that represents a "success"

b. Explain why x is (approximately) a binomial random variable

c. Give the value of p for this binomial experiment

d. Find P(x = 3)

e. Find the probability that 2 or fewer of the 10 U.S. adults do not work during summer vacation
Mathematics
1 answer:
Alex Ar [27]3 years ago
6 0

A becuase i worked it out and i got that so im really confident

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A circle has a circumference of 6. It has an arc of length 17/3. What is the central angle of the arc, in degrees?
zepelin [54]

Answer:

340 degrees

Step-by-step explanation:

So the key thing here is to notice that we are given the circumference which will allow us to find a value for the radius of the circle and hence the angle subtended by the arc (the central angle).

So the circumference of a circle = 2pi(r)

This means:

6 = 2pi(r)

Which means that

r = 6/2pi or r = 3/pi

Now we can use this value of r to find our angle in conjunction with the value of the arc length. So:

Arc length is defined by: length = θr

Where θ is our angle value.

So lets plug in:

\frac{17}{3} = (angle)\frac{3}{\pi }

Multiply by pi to get:

\frac{17\pi }{3} = 3(angle)

Divide by 3 to get that:

θ = 17pi/9

So if we convert that from radians to degrees we get 340 degrees.

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2 years ago
Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhjnrby
pishuonlain [190]

Answer:

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Step-by-step explanation:

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3 0
2 years ago
Please answer a through d
Elza [17]

Answer:

See explanation below.

Step-by-step explanation:

Given: 100 lbs on Earth is 16.6 lbs on the moon.

a. The independent variable is weight. The gravity of the Moon and the gravity of the Earth are constant. Weight can change, but gravity is a constant.

b. An equation that relates the weight of someone on the Moon who travels to the Earth:

100 / 16.6 = 6.02. Take the Moon weight and multiply by 6.02:

Moon Weight * 6.02 = Earth Weight.

Proof:

16.6 * 6.024 = 99.99 - approximately 100 lbs Earth weight.

c. A 185 lb astronaut on Earth would weigh:

16.6 / 100 = .166. Take the Earth weight and multiply by .166:

185 * .166 = 30 lbs on the Moon.

d. A person who weighs 50 lbs on the Moon:

50 * 6.024 = 301.2 lbs on Earth.

Hope this helps! Have a good day and year! :)

5 0
2 years ago
Kiera makes 181.5 ounces of punch for a pool party. She has 5 guests attending the pool party. How many ounces of punch does she
tresset_1 [31]
907.5 because you multiply 181.5 x 5 which equals 908.5
4 0
1 year ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

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

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
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