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mel-nik [20]
3 years ago
13

Help me plz i can't figure this out

Mathematics
1 answer:
zzz [600]3 years ago
6 0

Answer:

C

Step-by-step explanation:

For C the y axis changes, add 5 to -3 and you get 2. Therefore 5 units away!

Hope this helps, good luck! :)

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Estimate each product.Then write whether the estimate is greater than or less than the actual product.There are 62 rows of 9 cha
Westkost [7]
That is 62x9 which is 9x2=18 and 9x60=540 so you then add 540+18=558 chairs
8 0
3 years ago
Which expression shows how 6 • 45 can be rewritten using the distributive property?
tatuchka [14]
6*45=270 is the first step you should do to get the answer you are looking for. The next thing I would do is go to all the answers to see which one comes out the same way. The first one comes to be 246, the next one is 54, the next is 270 and the last one is 220. 6*40+6*5 is the correct answer.

4 0
3 years ago
A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi
Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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 normal distributions can be 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sum of n values of a distribution, the mean is \mu \times n and the standard deviation is \sigma\sqrt{n}

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that \mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300

If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload

Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

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

Z = \frac{20000 - 20000}{300}

Z = 0

Z = 0 has a pvalue of 0.5

1 - 0.5 = 0.5

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

5 0
2 years ago
(i) 809, 708, 607, _____, _ class 4<br>__class 4​
schepotkina [342]

Answer:

809, 708, 607, 506, 405, 304, 203, 102, 1, -100

Step-by-step explanation:

809, 708, 607, _____, _______

First term = 809

Second term = 708

Third term = 607

Difference between first term and second term = 809 - 708

= 101

Difference between second term and third term = 708 - 607

= 101

Therefore, the common difference is 101

Fourth term = 607 - 101

= 506

Fifth term = 506 - 101

= 405

Sixth term = 405 - 101

= 304

Seventh term = 304 - 101

= 203

Eighth term = 203 - 101

= 102

Ninth term = 102 - 101

= 1

Tenth term = 1 - 101

= - 100

809, 708, 607, 506, 405, 304, 203, 102, 1, -100

4 0
2 years ago
What is the 9th multiple of 15
MArishka [77]
135 because 
15x1=15 15x7=105
15x2=30 15x8=120
15x3=45 15x9=135
15x4=60 15x10=150
15x5=75 15x11=165
15x6=90 15x12=190

5 0
3 years ago
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