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3 years ago

Pls help this is rlly important!! You’ll get branliest bc this is hard and I’m stuck.

Mathematics

1 answer:

BaLLatris [955]3 years ago

4 0

the median of restaurant b's cleanliness ratings is 2.

the median of restaurant b's food quality ratings is 4.

the median of restaurant b's service ratings is 3.

:))

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A tree has circumference of 98 inches. Find the area of a circle with that circumference.

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Answer:

7542.96

Step-by-step explanation:

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2 years ago

A study shows that 70 out of 200 college students read the newspaper. Marysville College has 12,600 students.

Lubov Fominskaja [6]

Your answer would be 4,410

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3 years ago

Read 2 more answers

A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co

Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let $\mu$ = mean weight of coffee in its containers.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 32 ounces {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, $H_A$ : $\mu$ < 32 ounces {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = 31.8 ounces

s = sample standard deviation = 0.48 ounces

n = sample of containers = 15

So, the test statistics = $\frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }$ ~ $t_1_4$

= -1.614

The value of t test statistics is -1.614.

Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

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3 years ago

A group conducted a poll of 2083 likely voters just prior to an election. The results of the survey indicated that candidate A w

mart [117]

Answer:

With the given margin of error its is possible that candidate A wins and candidate B loses, and it is also possible that candidate B wins and candidate A loses. Therefore, the poll cannot predict the winner and this is why race was too close to call a winner.

Step-by-step explanation:

A group conducted a poll of 2083 likely voters.

The results of poll indicate candidate A would receive 47% of the popular vote and and candidate B would receive 44% of the popular vote.

The margin of error was reported to be 3%

So we are given two proportions;

A = 47%

B = 44%

Margin of Error = 3%

The margin of error shows by how many percentage points the results can deviate from the real proportion.

Case I:

A = 47% + 3% = 50%

B = 44% - 3% = 41%

Candidate A wins

Case II:

A = 47% - 3% = 44%

B = 44% + 3% = 47%

Candidate B wins

As you can see, with the given margin of error its is possible that candidate A wins and candidate B loses, and it is also possible that candidate B wins and candidate A loses. Therefore, the poll cannot predict the winner and this is why race was too close to call a winner.

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3 years ago

BRAINLIEST ANSWER!!!! #9 1/4 (5x+2)-1/5(x+3)=2 what’s the solution set?

Nina [5.8K]

x=2 from cross multiplying.

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3 years ago