What is the answer to y=-2/3+b

Divide. Write the quotient in lowest terms.<br> 5 divide 3 1/3

Step2247 [10]

Answer:

3/2 or 1 1/2

Step-by-step explanation:

5÷3 1/3

5/1 ÷ 10/3 because you have to get it into an improper fraction form in order to divide

5/1 × 3/10 because you have to multiply the first number by the reciprocal

5 and 10 cancel out so you would have 2 at the bottom

3/2 or 1 1/2

3 0

3 years ago

What is 10.5% of 238?

pshichka [43]

I believe that it is 24.99 but I'm not too sure. Anyways I hope I helped!

7 0

3 years ago

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DeSean is sorting his toy car collection. ⅔ of the cars are sports cars. ¼ of the remaining toys are pick up trucks. All of the

user100 [1]

Answer:

Fraction of fire trucks = $\frac{1}{4}$

Total number of toy cars = $16$

Step-by-step explanation:

Fraction of sports car = $\frac{2}{3}$

Fraction of remaining cars $=1-\frac{2}{3}=\frac{3-2}{3} =\frac{1}{3}$

So,

Fraction of pick up cars = $\frac{1}{4}(\frac{1}{3})=\frac{1}{12}$

Therefore,

Fraction of fire trucks = $\frac{1}{3}-\frac{1}{12}=\frac{4-1}{12}=\frac{3}{12}=\frac{1}{4}$

That is Number of fire trucks = $\frac{1}{4}$ ( Total number of toy cars )

Number of fire trucks = 4

Total number of toy cars = $4(4)=16$

3 0

3 years ago

Julie made $22 $55 $38 and $25 babysitting how much did she make babysitting

Nezavi [6.7K]

she made $140 dollars babysitting

hope it helps

8 0

3 years ago

Read 2 more answers

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?

$n = 2500$

So

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

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

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