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VARVARA [1.3K]
3 years ago
8

The standard normal curve shown below is a probability density curve for a

Mathematics
1 answer:
Step2247 [10]3 years ago
8 0

Answer:

0.6825

Explanation:

I'm not fully sure about how to do this, but I believe it might have to do with

the approx. amount of shade, but please take this explanation with a grain of salt as I'm not 100% certain. I am certain about the answer however.

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The manager at the concession stand offers drinks for sale on nine different prices. The major recorded the price each time in i
aleksandr82 [10.1K]

Answer:

20 drinks were sold, 3 were sold at the 5th price, and more were sold at a lower price.

Step-by-step explanation:

4 0
3 years ago
Solve the equation. Check your answer.<br><br> -11=5+8x<br><br><br> x=?
zloy xaker [14]

Answer:

x=-2

Step-by-step explanation:

Consider equation -11=5+8x

Collecting the like terms

-8x=11+5, -8x=16

Dividing through by -8 ob both sides of the equation

(-8x)-/8=16/-8

x=-2

4 0
3 years ago
HELP PLEASE EITHER ONE!! JUST SHOW ME HOW!! THANK YOU
Oksanka [162]
Try Socratic hope that helps
5 0
3 years ago
Steven was given the function, f(x)=1.05^2t and asked to determine the rate of change and the type of exponential function. What
erik [133]
f(x)=1.05^{2t}
The term 1.05 consists of the initial value 1 and the rate of change 0.05, (the latter being expressed as a decimal).
Therefore the correct answer choice is C.
6 0
3 years ago
What proportion of US women have a height greater than 69.5 inches?
kiruha [24]

Using the Normal distribution, it is found that 0.0359 = 3.59% of US women have a height greater than 69.5 inches.

In a <em>normal distribution</em> with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.

US women’s heights are normally distributed with mean 65 inches and standard deviation 2.5  inches, hence \mu = 65, \sigma = 2.5.

The proportion of US women that have a height greater than 69.5 inches is <u>1 subtracted by the p-value of Z when X = 69.5</u>, hence:

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

Z = \frac{69.5 - 65}{2.5}

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

1 - 0.9641 = 0.0359

0.0359 = 3.59% of US women have a height greater than 69.5 inches.

You can learn more about the Normal distribution at brainly.com/question/24663213

3 0
3 years ago
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