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Misha Larkins [42]
4 years ago
15

How high is a tree that casts a 2222​-ftft shadow at the same time a 44​-ftft postpost casts a shadow which is 77​-ftft ​long?

Mathematics
1 answer:
Zina [86]4 years ago
8 0

Answer: The height of the tree is 12.57 ft.

Step-by-step explanation:

The height of a thing is proportional to the its shadow.

Let H = Height and S = length of shadow

Then by equation direct proportion , \dfrac{H_1}{S_1}=\dfrac{H_2}{S_2}       (i)

Given : A tree that casts a 22 ft shadow at the same time a 4 ft postpost casts a shadow which is 7-ft ​long.

Put S_1=22 ,\ H_2= 4,\ \ S_2=7  in (i), we get

\dfrac{H_1}{22}=\dfrac{4}{7}\\\\ H_1=\dfrac{4}{7}\times22\approx12.57

Hence, the height of the tree is 12.57 ft.

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A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat
Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

Step-by-step explanation:

1) Data given and notation

s^2 =16 represent the sample variance

s=4 represent the sample standard deviation

\bar x represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

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 mean or variance 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.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=36.191

\chi^2_{1- \alpha/2}=7.633

And replacing into the formula for the interval we got:

\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}

8.400 \leq \sigma^2 \leq 39.827

So the 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

7 0
3 years ago
Assume the random variable x has a binomial distribution with the given probability of obtaining a success. Find the following.
borishaifa [10]

Answer:

P(x > 10) = 0.6981.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this question:

n = 14, p = 0.8

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P(x > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 11) = C_{14,11}.(0.8)^{11}.(0.2)^{3} = 0.2501

P(X = 12) = C_{14,12}.(0.8)^{12}.(0.2)^{2} = 0.2501

P(X = 13) = C_{14,13}.(0.8)^{13}.(0.2)^{1} = 0.1539

P(X = 14) = C_{14,14}.(0.8)^{14}.(0.2)^{0} = 0.0440

P(x > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.2501 + 0.2501 + 0.1539 + 0.0440 = 0.6981

So P(x > 10) = 0.6981.

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