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k0ka [10]
3 years ago
13

The trapezoid shown is divided into a right triangle and a rectangle. Write an expression that you could use to find area of the

trapezoid
The numbers are 6 in and 9 in btw
Mathematics
1 answer:
Aleks [24]3 years ago
5 0
I dont know the figure confuses me sorry
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The average speed of a golden eagle is 30 mph and the average speed of a peregrine falcon is 48 mph. How far will the peregrine
Oliga [24]

Answer:

53 1/3 miles.

Step-by-step explanation:

Well the eagle would take 1 1/3 of an hour to fly 40 miles, so you multiply 1 1/3 to 48 to get your answer.

5 0
2 years ago
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Find the midpoint of the line segment joining the points (7,1) and (-1,-1)
chubhunter [2.5K]
= x1 + x2/2 , y1 + y2/2
= 7+-1/2 , 1+(-1)/2
Midpoint = (3,0)
6 0
2 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
3 years ago
Question 5
Jobisdone [24]

Answer:just tryna get points don't mind me

Step-by-step explanation:

....

3 0
3 years ago
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?
Sloan [31]

Answer:

I got -16x+44/3

Step-by-step explanation:

8 0
2 years ago
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