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3 years ago

American Indian resistance in the Ohio River Valley ended with

Mathematics

2 answers:

Lisa [10]3 years ago

6 0

Answer:

A.) The Treaty of Greenville

Step-by-step explanation:

I hope the answer is correct, if not, I'm very sorry ♡

zmey [24]3 years ago

5 0

Answer:

American settlers and American Indians often clashed as settlement spread across the West during that era.

Step-by-step explanation:

American Indian resistance in the Ohio River Valley ended with.

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Why is 1/3 bigger than 4/5

UkoKoshka [18]

1/3 is in fact NOT larger than 4/5. 4/5 is larger than 1/3.

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3 years ago

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Can you help me out what is 8 times 604

skelet666 [1.2K]

The answer would be 4832

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3 years ago

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A class of 40 students took a test 20 students had an average score of 65. the other students had an average score of 52. what i

Korolek [52]

Answer:

One: 58.5

Two 36.6

Step-by-step explanation:

One

20 students * 65 = 1300 This is the total score of all 20 students. They may not all get 65, but 1300 is what their average total is.

20 students * 52 = 1040

The total score of all 40 students was 1300 + 1040 = 2340

The average for the entire class was 2340 / 40 = 58.5

Two

This question is done almost the same way.

Supposes the total number of members is 100. That means that 28 of them are men

28 * 33 = 924 This is the total age for the men

100 - 28 = 72 Out of 100, 72 of the people there are women.

72 * 38 = 2736

The total number of years of all of the members is

924 + 2736 = 3660

The average age of all members is 3660/100 = 36.6

Why did I pick 100 as the total membership? It is because a percentage is number out of 100. Am I allowed to invent things like that? Sure. I'm just using the definition of a percentage.

The average will make it come out the same.

7 0

2 years ago

A, B and C are events such that P(A) = 1/3, P(B) = ¼, and P(C) = 1/5. Find P(AUBUC) under each of the following assumptions: a)

marta [7]

(a) If $A,B,C$ are mutually exclusive, then

$P(A\cap B)=P(A\cap C)=P(B\cap C)=P(A\cap B\cap C)=0$

so we have

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}$

(b) If $A,B,C$ are mutually independent, then

$P(A\cap B)=P(A)P(B),$

$P(A\cap C)=P(A)P(C),$

$P(B\cap C)=P(B)P(C),$

$P(A\cap B\cap C)=P(A)P(B)P(C)$

so that

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)+P(A\cap B)+P(B\cap C))+P(A\cap B\cap C)$

$P(A\cup B\cup C)=\dfrac{47}{60}-\left(\dfrac1{12}+\dfrac1{15}+\dfrac1{20}\right)+\dfrac1{60}$

$P(A\cup B\cup C)=\dfrac35$

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3 years ago

Select all that apply

notsponge [240]

Answer:D

Step-by-step explanation:

2x + 3y = 12

-2x -2x

3y = -2x + 12

/3 /3 /3

Y = -2/3x + 4

6 0

3 years ago