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3 years ago

11

Eugene spent $14 on a magazine and some candy bars. if theagazine cost $6 and each candy bar cost $2 then hpw many candy bar did

he buy?

1 answer:

Inessa05 [86]3 years ago

5 0

14=6+2x
subtract 6 from both sides
8=2x
divide both sides by 2
x=4 candy bars

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2 years ago

Andre wants to buy a bike that has a price tag of $125. What was the wholesale cost of this bike?

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The bike store marks up the wholesale cost of all of the bikes they sell by 30%. Andre wants to buy a bike that has a price tag of $125. What was the wholesale cost of this bike?

Answer:

$96.15

Step-by-step explanation:

Let x be the wholesale cost of the bike.

Given:

Price tag of a bike = $125.

The bike store marks up the wholesale cost of all of the bikes they sell by 30%.

So we can find the wholesale cost of the bike by the following way.

$\frac{100+30}{100}\times x = 125$

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$x=\frac{125\times 100}{130}$

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Therefore, the wholesale cost of the bike is $96.15.

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2 years ago

Point N ( , 2 9) is on RS. A translation moves point N to its image, N'(6 3 , ). What is the distance, in units, between any poi

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent (Pew Research, June 5, 2014). An independe

enot [183]

Answer:

a)

$n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2$ , in which z is related to the confidence level.

b) A sample size of 991 is needed.

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 $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

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This is n for which $M = 0.03$ . So

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

$0.03 = z\sqrt{\frac{0.16*0.84}{n}}$

$0.03\sqrt{n} = z\sqrt{0.16*0.84}$

$\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}$

$(\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2$

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Question b:

99% confidence level,

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

$n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2$

$n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2$

$n = 990.2$

Rounding up

A sample size of 991 is needed.

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2 years ago