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3 years ago

13

On the shelves of Maria's bookcase,1/2 of the books are mysteries,1/10 are science fiction, 1/5 are biographies,and the remainin

g 4 are classics.How many books are in her bookcase?

1 answer:

PolarNik [594]3 years ago

7 0

There are 12 books in her bookcase

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Fraction 3/8 to percent?

arsen [322]

Answer:

37.5 percent or just 37

Step-by-step explanation:

You have to do 3 divided by 8 which gets you 0.375, once you do that, you move thr decimal to places to the right. Hopefully this helps.

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4 years ago

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Each leg of a 45-45-90 triangle has a length of 6 units what is the length of its hypotenuse

inysia [295]

Answer:

It's the option D

6 root 2 units

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4 years ago

Kara drove 36.9 miles before her car broke down. If Kara lives 45.55 miles away from work, how much further does she need to tra

Nitella [24]

Answer:

She needs to travel 8.65 miles

Step-by-step explanation:

We are given

Kara lives 45.55 miles away from work

So,

total distance is 45.55 miles

Kara drove 36.9 miles before her car broke down

so, distance traveled is 36.9 miles

now, we can find remaining distance

so,

remaining distance = ( total distance)-(distance traveled)

we can plug value

and we get

remaining distance = 45.55-36.9

remaining distance = 8.65 miles

So,

She needs to travel 8.65 miles

7 0

3 years ago

43 random movie theaters across the US were surveyed, and each reported its revenue (in $) from the film Gemini Man last weekend

pogonyaev

Answer:

a) For this case we can conclude with 95% of confidence that the true mean for the revenue (in $) from the film Gemini Man last weekend is between (5071.03;5748.24)

b) $df=n-1=43-1=42$

c) If we increase the confidence level the value for $t{\alpha/2}$ would increasse and then the margin of error would increase and the interval would be more wider. So then the best answer is:

A. The confidence interval would become wider

d) For this case the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we increase the sample size then we conclude that the interval would be narrower. So then the best answer is

B. The confidence interval would become narrower

e) The mean is given by:

$\bar X= \frac{5071.03+5748.24}{2}= 5409.64$

$ME= \frac{5748.24-5071.03}{2}= 338.61$

And the deviation can be founded :

$\sigma= \frac{ME \sqrt{n}}{t}= \frac{338.61* \sqrt{43}}{2.02} = 1099.198$

And then we can find the statistic:

$t = \frac{5000-5409.64}{\frac{1099.198}{\sqrt{43}}}= -2.44$

And the p value would be:

$p_v =1- P(t_{42}$

And the best anwer would be

A. The p-value would be greater than 0.1

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".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=43 represent the sample size

Calculate the confidence interval

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:

$df=n-1=43-1=42$

Since the confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,42)".And we see that $t_{\alpha/2}=2.02$ and this value is exactly the same for the normal standard distribution and makes sense since the sample size is large enough to approximate the t distribution with the normal standard distribution.

So on this case the 95% confidence interval is given by (5071.03;5748.24)

Part a

For this case we can conclude with 95% of confidence that the true mean for the revenue (in $) from the film Gemini Man last weekend is between (5071.03;5748.24)

Part b

$df=n-1=43-1=42$

Part c

If we increase the confidence level the value for $t{\alpha/2}$ would increasse and then the margin of error would increase and the interval would be more wider. So then the best answer is:

A. The confidence interval would become wider

Part d

For this case the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we increase the sample size then we conclude that the interval would be narrower. So then the best answer is

B. The confidence interval would become narrower

Part e

For this case we are testing $\mu >5000$

The mean is given by:

$\bar X= \frac{5071.03+5748.24}{2}= 5409.64$

$ME= \frac{5748.24-5071.03}{2}= 338.61$

And the deviation can be founded :

$\sigma= \frac{ME \sqrt{n}}{t}= \frac{338.61* \sqrt{43}}{2.02} = 1099.198$

And then we can find the statistic:

$t = \frac{5000-5409.64}{\frac{1099.198}{\sqrt{43}}}= -2.44$

And the p value would be:

$p_v =1- P(t_{42}$

And the best anwer would be

A. The p-value would be greater than 0.1

4 0

4 years ago

Captain Ishaan has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Luis and his merciless band of thieve

Sidana [21]

Answer: $\frac{4}{35}$

Step-by-step explanation: The probability of Captain Ishaan hitting the pirate's ship is:

P(A) = $\frac{4}{5}$

The probability of the pirate hitting Captain's ship is:

P(B) = $\frac{1}{7}$

So, the probability of both hitting each other is a probability of A and B occuring, which means:

P(A) . P(B) = $\frac{4}{5} . \frac{1}{7}$

P(A) . P(B) = $\frac{4}{35}$

The probability of both the pirate and the Captain hit each other's ship is

$\frac{4}{35}$ or 11.43%

3 0

3 years ago