The diagram shows a rectangular yard. Each small block is

Step-by-step explanation: There's 9 people over the age of 14. And 9 divided by the total number, 20, gives you 0.45. 0.45 multiplied by 100 is 45.

7 0

3 years ago

The sum of two numbers is 40 and their difference is 6.5, what is their product? show work please

Finger [1]

Let b>a,

a+b=40 so we can say a=40-b

We are told that b-a=6.5, using a found above in the equation gives us:

b-(40-b)=6.5

b-40+b=6.5

2b-40=6.5

2b=46.5

b=23.25, and since a=40-b

a=40-23.25

a=16.75

So a=16.75 and b=23.25

check...

16.75+23.25=40, 40=40

23.25-16.75=6.5, 6.5=6.5

4 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 $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$