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Lady bird [3.3K]

3 years ago

6

In a bag, there are 10 black plastic chips, 7 yellow, 5 red, and 3 green. Two chips are drawn at random with replacement. Find P

(black, then red).

1 answer:

ad-work [718]3 years ago

7 0

In a bag, there are 10 black plastic chips, 7 yellow, 5 red, and 3 green.

Total number of chips = 10+ 7 + 5 + 3 = 25

Probability = $\frac{Number \ of \ particular \ chips}{total \ number \ of \ chips}$

Probability ( back drawn) = $\frac{10}{25}}$

Two chips are drawn at random with replacement.. So black is replaced

Probability ( red drawn) = $\frac{5}{25}}$

Probability (black, then red) = $\frac{10}{25}}$ * $\frac{5}{25}}$

= $\frac{50}{625}}$ = $\frac{2}{25}}$

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Given that (x+2) is a factor of x^3+8, find the other quadratic factor.

Semmy [17]

If you know the formular a^3+b^3=(a+b)(a^2-ab+b^2), you can solve this problem.
8 is 2 cubed, so x^3+2^3=(x+2)(x^2-2x+4)
so the other quadratic factor is x^2-2x+4

3 0

2 years ago

The number of milligrams of a certain medicine a veterinarian gives to a dog varies directly with the weight of the dog. If the

Andre45 [30]

Answer:

Step-by-step explanation:

A direct variation equation is of the form

y = kx,

where, in words, it reads "y varies directly with x" or "y varies directly as x". In order to use this as a model, we have to have enough information to solve for k, the constant of variation. The constant of variation is kind of like the slope in a straight line. It rises or falls at a steady level; it is the rate of change.

We have that a vet gives a dose of three-fifths mg to a 30 pound dog. If the dose varies directly with the weight of the dog, then our equation is

d = kw and we need to find k in order to have the model for dosing the animals.

$\frac{3}{5} =k(30)$

Divide both sides by 1/30 to get k alone.

$(\frac{1}{30})(\frac{3}{5})=k$ and

$k=\frac{1}{50}$

Our model then is

$d=\frac{1}{50}w$

This means that for every pound of weight, the dog will get one-fiftieth of a mg of medicine.

7 0

3 years ago

Read 2 more answers

Gary and Jill both want 23 stickers for a school project there are 3 sheets of 10 and 30 single stickers on the table how could

Ray Of Light [21]

Answer:

See below.

Step-by-step explanation:

There are three sheets of 10 stickers. This is equal to 30 stickers. There are also 30 single stickers. Split the single stickers and give Gary and Jill 13 each. Then Gary and Jill can each take a sheet of 10. They will both have 23 stickers.

7 0

3 years ago

A. In a school year the ratio of boys to girls is 3:4. If there are 372 boys, how many girls are there

natima [27]

Answer:

a. 496 girls

b. 110 boys

Step-by-step explanation:

a. 3:4 ------- divide 372 by 3 and you get 124 now multiply that by 4 and you get 496

b. 2:5 -------- divide 275 by 5 and you get 55 now multiply that by 2 and you get 110

7 0

3 years ago

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a <em>sample mean</em>. The distribution for <em>sample means</em> follows a <em>normal distribution</em> with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable <em>z</em> follows a <em>standard normal distribution</em>, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a <em>standard deviation</em> of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a <em>cumulative standard normal table</em> (or use some statistic program) to find the cumulative probability for <em>z</em> (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

5 0

3 years ago