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3 years ago

PLEASE HELP TIMED

Mathematics

1 answer:

rusak2 [61]3 years ago

5 0

A - drawing a red 8

B - drawing an 8

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}\\\\ |\Omega|=52\\ |A\cap B|=2\\ |B|=4\\\\ P(A \cap B)=\dfrac{2}{52}=\dfrac{1}{26}\\ P(B)=\dfrac{4}{52}=\dfrac{1}{13}\\\\ P(A|B)=\dfrac{\dfrac{1}{26}}{\dfrac{1}{13}}=\dfrac{1}{26}\cdot13=\dfrac{1}{2}$

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What 78 divide by 858 do yall know

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0.0909090909 Is the answer my friend plz mark brainliest

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3 years ago

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Jason jumped off of a cliff into the ocean in Acapulco while vacationing with some friends. His height above ocean measured in f

lbvjy [14]

Answer:

480 feet.

Step-by-step explanation:

We are told that the function $h(t)=-16t^2+16t+480$ models Jason's height above ocean measured in feet as a function of time and t is the time in seconds from jumping off.

To find the height of cliff we need to substitute t=0 in our given function as at t=0 we will get Jason's height above ocean which is same as the height of the cliff.

Upon substituting t=0 in our function we will get,

$h(0)=-16(0)^2+16*0+480$

$h(0)=-0+0+480$

$h(0)=480$

Since, the function gives Jason's height above ocean in feet, therefore, the cliff was 480 feet high.

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3 years ago

If we list all the possible factors (in order) of 12, we get 1, 2, 3, 4, 6, 12. This is the case because 1 x 12, 2 x 6, and 3 x

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Answer:

1, 2, 3, 5, 6, 10, 15 and 30.

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3 years ago

Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the

Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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.

In this problem, we have that:

$\mu = 9.6, \sigma = 0.5$ .

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when $X = 10$ .

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

$Z = \frac{10 - 9.6}{0.5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

$Z = \frac{8 - 9.6}{0.5}$

$Z = -3.2$

$Z = -3.2$ has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have $n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1$ .

This probability is 1 subtracted by the pvalue of Z when $X = 9.8$ . So:

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

$Z = \frac{9.8 - 9.6}{0.1}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

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3 years ago

Select the correct answer. The volume of one of the Great Lakes is 3. 5 × 103 cubic kilometers. If there are 6. 3 × 107 fish in

valentinak56 [21]

The average number of fish per cubic kilometer is $1. 8 \times 10^4$ .

The correct option is B.

Given

The volume of one of the Great Lakes is 3.5 × 103 cubic kilometers.

The number of fish in the lake = 6. 3 × 107.

<h3>What is the average value?</h3>

The average is the middle value of the given data set and is calculated by adding up all the numbers and dividing the result by the total quantity of figures added.

The average number of fish per cubic kilometer is given by;

$\rm \text{The average number of fish per cubic kilometer}= \dfrac{Number \ of \ fish \ in \ lake}{Total \ volume \ of \ lake}$

Substitute all the values in the formula;

$\rm \text{The average number of fish per cubic kilometer}= \dfrac{Number \ of \ fish \ in \ lake}{Total \ volume \ of \ lake}\\\\ \text{The average number of fish per cubic kilometer}= \dfrac{6.3\times 10^7}{3.5\times 10^3}\\ \text{The average number of fish per cubic kilometer}= 1.8 \times 10^4$

Hence, the average number of fish per cubic kilometer is $1. 8 \times 10^4$ .

To know more about the Average rate click the link given below.

brainly.com/question/12395856

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2 years ago