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Volgvan
2 years ago
12

5

Mathematics
1 answer:
Bezzdna [24]2 years ago
6 0

Answer:

120

Step-by-step explanation:

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The price of a 2-pack shopkin should not be more then 4$. choose an inequality for this situation
lutik1710 [3]
The price of the shopkin should be smaller or or equal


Solution: Should not be more than 4


Answer: <4
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5 0
2 years ago
There are 14 teams in a basketball league.
Talja [164]

Answer:

Step-by-step explanation:

Given that :

Number of teams = 14

Each team plays every other team twice ;

Using the combination formula :

nCr = n! ÷ (n-r)! r!

14C2 = 14! ÷ (14 - 2)! 2!

14C2 = 14! ÷ (12)! 2!

14C2 = (14 * 13) ÷ 2 * 1

14C2 = 182 / 2

14C2 = 91

Hence, since they are going to be playing each other twice :

2(14C2)

2 * 91 = 182games

7 0
3 years ago
Is this the right answer?
Ne4ueva [31]
Yes your answer is solved correctly
6 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
3 years ago
24+82 rounding or compatible numbers to estimate the sum
allsm [11]

Answer:

106

Step-by-step explanation:

you take 82 and 24 and add them together so it would be like this 2 + 4 = 6 and 8 + 2 = 10 so the 8 is in a tens place so the answer from adding 8 + 2 = 100 because you in the tens place so the final answer would be 106 after adding it all up.

4 0
2 years ago
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