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Lemur [1.5K]
2 years ago
14

An artist uses 18 colors on an art palette to paint a picture. This is 72% of all the colors on the palette. How many colors are

on the artist’s palette?
Mathematics
2 answers:
alekssr [168]2 years ago
4 0

Answer:

25

Step-by-step explanation:

Think the number is x

So

According to the question

The equation is

x×72%=18

x×72/100=18

x×18/25=18

x=18×25/18

x=25

So

There are 25 colours on the artist's palette

brilliants [131]2 years ago
3 0

Answer:

He actually has 25 paints.

Step-by-step explanation:

This is because you take 18x100(100 is the total % under 72 because 72 is out of 100) that equals 1800. then, you take 1800 and divide it by 72. That is because 72 is the actual number that we already have. You would get 25 after dividing. hope this helps:)

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Read 2 more answers
Reading proficiency: An educator wants to construct a 99.5% confidence interval for the proportion of elementary school children
Shkiper50 [21]

Answer:

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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 of the interval is given by:

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

In this problem, we have that:

p = 0.69

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

This is n when M = 0.07. So

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

0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}

0.07\sqrt{n} = 1.2996

\sqrt{n} = \frac{1.2996}{0.07}

\sqrt{n} = 18.5658

(\sqrt{n})^{2} = (18.5658)^{2}

n = 345

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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