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vaieri [72.5K]
3 years ago
7

the largest elephant weighs 30.25 times as much as the largest gorillas so if the largest elephant weighs 14,520 pounds how much

do the largest gorillas weigh
Mathematics
2 answers:
pav-90 [236]3 years ago
6 0

Let the elephant = e

Let the gorillas = g

e = 30.25 * g

e = 14,520lbs

(14,520lbs)/30.25=(30.25)/30.25*g

480lbs=g

<h2>The largest gorillas weigh 480 lbs.</h2>

ok so basically im monkey

Aleks04 [339]3 years ago
5 0

Since the elephant weighs 14,520 pounds, and weighs 30.25 times as much as the largest gorilla. You would divide 14,520 by 30.25 to get 480. The largest gorilla is 480 pounds.

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liraira [26]

Answer:

11

Step-by-step explanation:

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3 years ago
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1. A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts o
Sauron [17]

Answer:

The confidence interval estimate of the population mean is :

(0.61 ppm, 0.90 ppm)

The correct option is (A).

Step-by-step explanation:

The amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city are:

S = {0.58, 0.82, 0.10, 0.98, 1.27, 0.56, 0.96}

A (100-\alpha )\% confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a (100-\alpha )\% probability of consisting the true value of mean.

⇒ Since the population standard deviation is not provided we will use the <em>t</em>-distribution to construct the 99% confidence interval for mean.

⇒ The formula for confidence interval for the population mean is:

                                          \bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}

Here,

\bar x = sample mean

<em>s </em>= sample standard deviation

<em>n</em> = sample size

t_{\alpha/2,(n-1)} = critical value.

The degrees of freedom for the critical value is, (<em>n</em> - 1) = 7 - 1 = 6.

The significance level is: \alpha =1-Confidence\ level=1-0.99=0.01

The critical value is:

t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143

**Use the <em>t</em>-table for the critical value.

Compute the sample mean and sample standard deviation as follows:

\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753

\int\limits^a_b {x} \, dx s=\sqrt{\frac{\sum (x{i}-\bar x)^{2}}{n-1} } =\sqrt{\frac{1}{6} \times 0.859743} =0.379

The 99% confidence interval for μ is:

x^{2} CI=0.753\pm 3.143\times\frac{0.379}{\sqrt{7}} \\=0.753\pm0.143\\=(0.61, 0.896)\\\approx(0.61, 0.90)

The confidence interval estimate of the population mean is:

(0.61 ppm, 0.90 ppm)

The upper and lower limit of the 99% confidence interval indicates that the true mean value is less than 1 ppm. This implies that there is not too much mercury in tuna sushi

Because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

Thus, the correct option is (A).

6 0
3 years ago
An airport is studying the noise levels of jets during takeoff as they pass over a neighborhood. They find that at the present t
avanturin [10]

Answer:  2.28%

Step-by-step explanation:

Given: An airport find that at the present time, the mean noise level is 103 decibels and the standard deviation is 5.4 decibels.

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The probability that jets have a noise level less than 92.2 decibels when taking off over this neighborhood will be :-

P(X

Hence, the percent of these jets have a noise level less than 92.2 decibels when taking off over this neighborhood = 2.28%

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3 years ago
How much will it cost to ride if the cab travels 13.5 miles?
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ExtremeBDS [4]

we are given

\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}

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we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

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so, one of restriction is

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we can simplify it

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we know that denominator can not be zero

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so, option-B.......Answer

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3 years ago
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