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3 years ago

ACTIVITY 1

Mathematics

1 answer:

Shkiper50 [21]3 years ago

6 0

Answer:

500,000

(3)

10.5%

Activity 2

Directions: Find the unknown principal P, maturity

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Why does fast- moving grader erosion and then slow- moving water?

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It have more energy picking up more particles in less time

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3 years ago

Angle ACB is an inscribed angle of circle P. What is the measure of angle ACB if arc MAB is equal to 50 degrees?

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∠ACB is an inscribed angle, so

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3 years ago

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A college conducts a common test for all the students. For the Mathematics portion of this test, the scores are normally distrib

Jet001 [13]

Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 502, \sigma = 115$

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:

X = 590:

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

$Z = \frac{590 - 502}{115}$

Z = 0.76

Z = 0.76 has a p-value of 0.7764.

X = 400:

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

$Z = \frac{400 - 502}{115}$

Z = -0.89

Z = -0.89 has a p-value of 0.1867.

0.7764 - 0.1867 = 0.5897 = 58.97%.

58.97% of students would be expected to score between 400 and 590.

More can be learned about the normal distribution at brainly.com/question/27643290

#SPJ1

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1 year ago

How many ways can you arrange the letters in the word prime?

Jobisdone [24]

There are 5 letters in the word "prime"

Imagine we had 5 slots to fill. They are empty initially.
Slot 1 has 5 choices to pick from
Once we pick a letter, we have 4 choices left over for slot 2
Slot 3 will have 3 choices
Slot 4 will have 2 choices
Slot 5 will have 1 choice

We have this countdown: 5,4,3,2,1

which multiplies out to 5*4*3*2*1 = 120

There are 120 unique ways to arrange the letters. Order matters. Because order matters, this is a permutation.

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3 years ago

Ten pieces of paper numbered consecutively from 1 to 10 are put into a hat. if three pieces of paper are drawn at random from th

DaniilM [7]

There are 5 even numbers from 1 to 10.

The number of ways to draw all even numbers is 5!/(2!*3!) = 10

The number of ways to draw from the hat is 10!/(7!*3!) = 128

Therefore the probability is 10/128 = 5/64 or roughly 7.81%

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2 years ago