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asambeis [7]
3 years ago
9

On average, who spends more money on Prom? Forty-two senior girls and thirty-nine senior boys were asked how much they plan on s

pending on prom this spring.
1 proportion z-test

2 proportion z-test

1 sample t-test

2 sample t-test

Paired t-test

1 proportion z-interval

2 proportion z-interval

1 sample t-interval

2 sample t-interval

Paired t-interval

?
Mathematics
1 answer:
blagie [28]3 years ago
4 0

Answer:

Yo I think we're in the same class.... wish i knew the answer lol

Step-by-step explanation:

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They Y intercept is 1
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Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function: y=4x^2-8x-3
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D. x= -1; vertex: (-1,-7) 
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(FIRST PERSON TO ANSWER GETS BRAINLIEST!!!) A student earned $2500.75 at his summer job making $12.50 per hour. Let h represent
laiz [17]

<u>Answer:</u>

12.50 \space\ h = 2500.75

<u>Step-by-step explanation:</u>

We know from the question that the student earned $12.50 <em>per hour</em>.

Using this information, we can say that if the student worked for <em>h </em>hours, they would make a total of 12.50 × <em>h </em>dollars.

We also know that the total money they earned is $2500.75.

∴ Therefore, we can set up the following equation:

\boxed {12.50 \times h = 2500.75}

From here, if we want to, we can find the number of hours worked by simply making <em>h</em> the subject of the equation and evaluating:

<em>h </em>=<em> </em>\frac{2500.75}{12.50}

  = 200.6 hours

5 0
2 years ago
Read 2 more answers
A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimat
Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

For this problem, we consider that we want it to be within 4%.

Item a:

  • The sample size is <u>n for which M = 0.04.</u>
  • There is no estimate, hence \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2

n = 600.25

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is \pi = 0.96, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}

\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2

n = 92.2

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to \pi = 0.5, the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151  

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2 years ago
Two less than four times a number is 6
iris [78.8K]

Answer: 2 - 4x = 6

Please mark me as brainliest

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