Which expressions are equivalent to -90-60w? Choose 3 answers:

In a school of 330 students, 85 of them are in a drama club, 200 of them are in a sports team and 60 of them do both

Neko [114]

Answer:

105/330 = 21/66 = 7/22

7/22 is the answer I think

Step-by-step explanation:

n(U) = 330

n(A) = 85

n(B) = 200

n(A n B) = 60

Now,

n(A U B) = n(A) + n(B) - n(A n B)

or, n(A U B) = 85 + 200 - 60

or, n(A U B) = 285 - 60 = 225

Now,

n(A U B) compliment = n(U) - n(A U B)

n(A U B) compliment = 330 - 225

so, n(A U B) compliment = 105

Now,

probability = 105/330

8 0

2 years ago

6^2 times 3/2 dived by 71/54 plz actually help<br><br> anyone got pad let

algol [13]

Answer:

41.0704225352 or 284x/9 + 0

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

4/5x+5=9 solve for x please i have 8 questions left!!!!!!!!!!!

NemiM [27]

Answer:

x= 5

Step-by-step explanation:

You need to get rid of the fraction first and to do that you muliply the whole equation by 5 so that your new equation reads 4x+25=45. Then you subtract 25 from both sides so you have 4x=20. Divide both sides by 4 so x=5.

5 0

3 years ago

Maureen rolls a fair dice and flips a fair coin.

Taya2010 [7]

1/6 times 1/2 = 1/12

5 0

2 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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 normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$