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Exercise 1: In the right triangle below, calculate the measurement of the unknown side. Show your work.

vekshin1

Answer:

Exercise 1:

base [b]=8cm

perpendicular [p]=6cm

hypotenuse [h]=?

By using Pythagoras law

h²=p²+b²

h²=6²+8²

h=√100

h=10cm

So another side's length is 10cm

Exercise 2:

base [b]=6m

perpendicular [p]=bm

hypotenuse [h]=8m

By using Pythagoras law

h²=p²+b²

8²=b²+6²

b²=8²-6²

b=√28=2√7 0r 5.29 or 5.3

So height of kite is√28or 2√7 0r 5.29 or 5.3 m

Step-by-step explanation:

[Note: thanks for translating]

4 0

3 years ago

What is the radical expression

Margaret [11]

$5x^{\frac{1}{4} }=5\cdot x^{\frac{1}{4}}= 5\sqrt[4]{x} \\\\because\\\\a^{\frac{1}{n} }=\sqrt[n]{a}$

6 0

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