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garik1379 [7]
3 years ago
6

The following data points represent the number of jars of honey Martha the Bear consumed each day this week. \qquad4,4,5,2,2,3,

44,4,5,2,2,3,44, comma, 4, comma, 5, comma, 2, comma, 2, comma, 3, comma, 4 Using this data, create a frequency table. Number of jars of honey Number of days 222 333 444 55
Mathematics
1 answer:
ikadub [295]3 years ago
3 0

What are the commas for?  I think this was copied wrong, but I'd say use the # of bears eaten/jars as the (y) variable and the # of days as the (x) variable.

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The probability that a certain hockey team will win any given game is 0.3723 based on their 13 year win history of 385 wins out
Whitepunk [10]

Answer:

0.1505 = 15.05% probability that the hockey team wins 6 games in November

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the team wins, or it does not. The probability of winning a game is independent of winning other games. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability that a certain hockey team will win any given game is 0.3723

So p = 0.3723

12 games in November

So n = 12

What is the probability that the hockey team wins 6 games in November?

This is P(X = 6)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{12,6}.(0.3723)^{6}.(1-0.3723)^{6} = 0.1505

0.1505 = 15.05% probability that the hockey team wins 6 games in November

4 0
3 years ago
A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If
musickatia [10]

Answer:

a. The test statistic for this hypothesis would be z=1.71

b. Critical value at α = 0.10: z=1.282

c. Reject H0, the television network should not keep its current lineup.

d. H0: p ≤ 0.50; HA: p > 0.50

Step-by-step explanation:

We have to performa an hypothesis test on a proportion.

The claim is that the proportion of viewers that prefer the new show is bigger than 50% (meaning that most viewers prefer the new show).

The null hypothesis will state that both shows have the same proportion.

Then, the null and alternative hypothesis are:

H_0: \pi\leq0.50\\\\H_a: \pi>0.5

The sample proportion is p=0.53

p=X/n=438/827=0.53

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.5*0.5}{827}}= 0.017

Then, the z-statistic can be calculated as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.53-0.50-0.5/827}{0.017}=\dfrac{ 0.029 }{0.017} =1.706

The critical value for a right tailed test at a significance level of 0.10 is zc=1.282. This value can be looked up in a standard normal distribution table.

As the statistic is bigger than the critical value, it lies in the rejection region. The null hypothesis is rejected. There is evidence to support the claim that the proportion of viewers which support the new show is larger than 0.50.

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Answer:

Step-by-step explanation:

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Tell whether each statement is True or False.
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Answer:

3+5(9+2)= 48+5b true true

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Step-by-step explanation:

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