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2 years ago

Somebody please help me solve this I’m stuck

Mathematics

1 answer:

madam [21]2 years ago

7 0

7. 2; 8. 1/4; 10. -25; 12: -3
(remember to use rise over run method and notice the position of each linear slope)
hopefully that helped! <3

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Heeeeeeeeeelpppppppp meeeeeeee

Step2247 [10]

Answer: The answer is 10.5

Step-by-step explanation: for subtracting the answer is 1.5 and for multplying its 39.375 and last for division its 0.51428571428 *lol*

Have a wonderful day! ^-^

6 0

3 years ago

A report on consumer financial literacy summarized data from a representative sample of 1,664 adult Americans. Based on data fro

daser333 [38]

Answer:

a. The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

b. Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Sample of 1,664 adult Americans, 939 people in the sample would have given themselves a grade of A or B in personal finance.

This means that $n = 1664, \pi = \frac{939}{1664} = 0.5643$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 - 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.54$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 + 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.588$

The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

(b) Is the confidence interval from part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of A or B?

Yes, because the confidence interval is entirely above 0.5.

Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

8 0

3 years ago

On Monday, a museum had 600 visitors. On Tuesday, it had 740 visitors. Estimate the percent change in the number of visitors to

prisoha [69]

15 percent I think :)))))))))))))))))(

3 0

3 years ago

A parabola intersects the xxx-axis at x=3x=3x, equals, 3 and x=9x=9x, equals, 9.

vekshin1

Answer:

$x^2-12x+27 =0$