How do you write an equation 18 more than the product of 2 and 42

Jada has some pennies and dimes. The ratio of Jada's pennies to dimes is 2 to 3. From the information given, can you determine h

Anuta_ua [19.1K]

Answer:

No, we can't determine how many coins Jada has.

She will have 22 pennies and 33 dimes.

4 0

3 years ago

Two angles are complementary. The measure of one angle is 3° more than one-half of the measure of the other. Find the measure of

vagabundo [1.1K]

Answer:

doge

Step-by-step explanation:

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8 0

3 years ago

Sole for |p+8| over 2 =5

Lubov Fominskaja [6]

Answer:

p =2 p=-18

Step-by-step explanation:

|p+8|

-------------- = 5

2

Multiply each side by 2

|p+8|

-------------*2 = 5*2

2

|p+8| =10

There is a positive and a negative solution

p+8=10 p+8=-10

Subtract 8 from each side

p+8-8=10-8 p+8-8=-10-8

p =2 p=-18

7 0

4 years ago

HELPPPLPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

MAXImum [283]

C May be the answer sorry if it is wrong.

5 0

3 years ago

Read 2 more answers

6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that woul

drek231 [11]

Answer:

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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 of:

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

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that $n = 1012, \pi = \frac{374}{1012} = 0.3696$

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 $Z = 1.96$ .

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

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

$E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}$

$E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}$

$\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

$n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

4 0

3 years ago