PLEASEEE HELP ILL GIVE BRAINLIST what is 21+3???? PLEASEEEEEE

A museum curator is hanging 7 paintings in a row for an exhibit. There are 4 Renaissance paintings and 3 Baroque paintings. From

Marina CMI [18]

Answer:

144 ways

Step-by-step explanation:

Number of paintings = 7

Renaissance = 4

Baroque = 3

We are hanging from left to right and we will first hang Renaissance painting before baroque painting.

For Renaissance we have 4! Ways of doing so. 4 x3x2x1 = 24

For baroque we have 3! Ways of doing so. 3x2x1 = 6

We have 4!ways x 3!ways

= (4x3x2x1) * (3x2x1) ways

= 144 ways

Therefore we have 144 ways to hang the painting.

6 0

3 years ago

PLEASE HELP! Y=5x-9 -2x-3y=-7 Please show work! substition for math

Zepler [3.9K]

Answer:

ill edit this and Post the answer in a few m0ments .....

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Ashley ran from home to school in 10 minutes. What is her average speed, if the distance between her house and her school is 1.5

wel

(.15 miles / 1 minute) * (60 minutes / 1 hour) = 9 miles / hour

6 0

3 years ago

Can someone help me??

Mariana [72]

Answer:

24 i think if im wrong im sorry :(

Step-by-step explanation:

3 0

2 years ago

A certain brand of candies have a mean weight of 0.8616g and a standard deviation of 0.0518 based on the sample of a package con

torisob [31]

Answer:

a) The probability is 0.5557 = 55.57%.

b) The probability that a sample of 447 candies will have a mean of 0.8542g or greater is 0.9987 = 99.87%.

c) Yes, because there is a very large probability, of 99.87%, that the amount will be at least the one claimed on the label.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean weight of 0.8616g and a standard deviation of 0.0518

This means that $\mu = 0.8616, \sigma = 0.0518$

a) If 1 candy is reandomly selected, find the probability that it weights more than 0.8542g.

This is 1 subtracted by the pvalue of Z when X = 0.8542. So

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

$Z = \frac{0.8542 - 0.8616}{0.0518}$

$Z = -0.14$

$Z = -0.14$ has a pvalue of 0.4443

1 - 0.4443 = 0.5557

The probability is 0.5557 = 55.57%.

b) If 447 candies are reandomly selected find the probability that their mean weight is at least 0.8542 g.

Sample of 447 means that $n = 447, s = \frac{0.0518}{\sqrt{447}} = 0.00245$

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

By the Central Limit Theorem

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

$Z = \frac{0.8542 - 0.8616}{0.0245}$

$Z = -3.02$

$Z = -3.02$ has a pvalue of 0.0013

1 - 0.0013 = 0.9987

The probability that a sample of 447 candies will have a mean of 0.8542g or greater is 0.9987 = 99.87%.

c) Given these results does it seem that the candy company is providing consumers with the amount claimed on the label?

Yes, because there is a very large probability, of 99.87%, that the amount will be at least the one claimed on the label.

6 0

2 years ago