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Andrej [43]
3 years ago
15

You are on a hockey team. Your team on the ice is 12 minutes per game. A season is twenty games. How many minutes do u play duri

ng the season
Mathematics
2 answers:
Lesechka [4]3 years ago
8 0
240 minutes per season
professor190 [17]3 years ago
6 0
240minutes per season
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The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

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.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

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.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
3 years ago
Please Help!! Give Brainly!!<br> What is the value of (14^0) ^-2?<br> -1/196<br> 1/196<br> 0<br> 1
Trava [24]

Answer:

so option 1 is correct

Step-by-step explanation:

(14^0) ^-2

power 0 * -2 = 0

if power is 0 then it equals to 1  so

14^0

= 1

8 0
2 years ago
What are natural numbers?
scoundrel [369]
Natural numbers are lie 1 2 3 4 5 6 7 8 9 and so on
8 0
3 years ago
Read 2 more answers
Please help me with these questions
Citrus2011 [14]

Answer:

okay the first questions answer is ,c,is true and rest of them are false

8 0
3 years ago
A college student is taking two courses. The probability she passes the first course is 0.67. The probability she passes the sec
andrezito [222]

Answer:

0.58 = 58% probability she passes both courses

Step-by-step explanation:

We can solve this question treating the probabilities as a Venn set.

I am going to say that:

Event A: She passes the first course.

Event B: She passes the second course.

The probability she passes the first course is 0.67.

This means that P(A) = 0.67

The probability she passes the second course is 0.7.

This means that P(B) = 0.7

The probability she passes at least one of the courses is 0.79.

This means that P(A \cup B) = 0.79

a. What is the probability she passes both courses

This is P(A \cap B).

We use the following relation:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

So

P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.67 + 0.7 - 0.79 = 0.58

0.58 = 58% probability she passes both courses

5 0
3 years ago
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