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3 years ago

Joey wants to collect 20 coins from foreign countries. If he has 3 coins so far, what percent of his collection is complete?

Mathematics

2 answers:

irinina [24]3 years ago

8 0

Answer:

17%

Step-by-step explanation:

I think you maybe have to subtract

Sphinxa [80]3 years ago

8 0

Answer:

15%

hope this helps!

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Line segment CD is shown on a coordinate grid:

Jet001 [13]

A) would be your answer hope this helps. :)

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3 years ago

Read 2 more answers

What is the equation for (4, 9.4) and (-3, -9.5)

Kipish [7]

Answer: I got an eqaution I hope it is basic

Step-by-step explanation:

4 times 9

9 - (-3 ) times -9 + 5

8 0

3 years ago

Which is equivalent to (2^-3)^4 1/2^12 2 -2^12 1/2^7

sleet_krkn [62]

Answer:

The answer is First option

1/2^12 .

$\frac{1}{2^{12} }$

Step-by-step explanation:

Given:

(2^-3)^4

i.e

$(2^{-3} )^{4}$

To Find:

$(2^{-3} )^{4}=?$

Solution:

Law of Indices

$(x^{a}) ^{b} =x^{a\times b}\\and\\x^{-1}=\frac{1}{x}$

Using the above Property we get

∴ $(2^{-3} )^{4}=2^{-3\times 4} \\\\ (2^{-3} )^{4}=2^{-12} \\\\(2^{-3} )^{4}= \frac{1}{2^{12} }$

The answer is First option

1/2^12 .

$(2^{-3} )^{4}=\frac{1}{2^{12} }$

6 0

3 years ago

Brandon's Crackers will make 8,648 ounces of cheese crackers next year. The company plans to put the crackers into 47-ounce boxe

Murljashka [212]

Answer:

184 boxes

Step-by-step explanation:

8648/47 = 184

8 0

2 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2500 individ

Ne4ueva [31]

Answer:

The approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they carry the defective gene that causes inherited colon cancer, or they do not. The probability of a person carrying this gene is independent from other people. So the binomial probability distribution is used to solve this question.

A sample of 2500 individuals is quite large, so we use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer.

This means that $p = \frac{1}{500} = 0.002$

In a sample of 2500 individuals, what is the approximate distribution of the number who carry this gene?

$n = 2500$

$\mu = E(X) = np = 2500*0.002 = 5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.002*0.998} = 2.23$

So the approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

4 0

3 years ago