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Yakvenalex [24]
3 years ago
9

!!!3 QUESTIONS!!!

Mathematics
1 answer:
RUDIKE [14]3 years ago
6 0

Answer:

1.) He still owes $100 of debt

2.) 1 rose would be .75 cents and 1 carnation would be .90 cents

3.) C 8 and 3/8

Step by Step explanation

(/ means divided by in the first 2 equations but / in the last question is for the fractions.)

1.) _(1,500 / 3 =500. 1,500 - 500 =1,000 / 5 =200 x 3 =600. 700 - 600 =$100)

2.)_(1.50/2 =.75 cents. 2.70/3 =.90 cents.)

3.) i first added the fractions ( i equalized them) then i added the whole numbers (the u in the equation was to separate the whole num. and fraction )

Before i equalized it: (6u3/4 + 1u1/8 = 7u1/8 + 1/8 = 8u3/8)

After i equalized it: (6u6/8 + 1u1/8 = 7u1/8 + 1/8 = 8u3/8)

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Statistics: Hypothesis Testing
Lina20 [59]

Answer:

Yes, the mean age of onset of anorexia nervosa in women has decreased.

Step-by-step explanation:

When constructing confidence intervals, remember the acronym PANIC.

P (p definitions)

let p = the newly recorded mean age of the onset of anorexia nervosa in women

A (assumptions)

There is no evidence in the problem that the women were randomly or independently selected, but we will proceed as if they were. It is reasonable to assume that there are over 200 women in the world, so this sample is less than 10% of the total population. We will assume that the onset of anorexia nervosa has a fairly normal distribution, so the smaller sample size will suffice.

N (name the test)

Because the required conditions are met, we can construct a 95% confidence interval.

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Interval = \overline{x} \pm Z^{*} (\frac{\sigma}{\sqrt{n} })

Interval = 14.250 \pm 1.96 (\frac{1.512}{\sqrt{20} })

Interval = 14.250 \pm .6626

Interval = (13.5874, 14.9126)

C (conclusion)

Based on this interval, I am 95% confident that the true mean age of the onset of anorexia nervosa in women is between 13.58 and 14.91 years. Since 15 is not included in this interval, I believe that the mean age of the onset of anorexia nervosa in women has decreased.

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