MATHEMATICS COLLEGE

-1 + 0 + 1 + ... + 12

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

-1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 77

Hope this helps :)

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Which graph shows the solution to the system of linear inequalities? Y< 1/3x-1 y<1/3-31) All values that satisfy y<1/3-1 are solutions
2) All values that satisfy y<1/3-3 are solutions
3) All values that satisfy either equations are solutions
4) There are no solutions
(Edge 2020)

Answers

The solution of the system of linear inequalities is All values that satisfy y<1/3-3 are solutions. (option 2)

What is inequality?

A relationship between two expressions or values that are not equal to each other is called inequality.

Given is a graph of inequalities, y < 1/3x-1 and y < 1/3-3

The solution of the inequality A is the shaded area below the solid blue line and the solution of the inequality B is the shaded area below the solid red line,

That means all the solutions of inequality B will satisfy the inequality A also.

Hence, The solution of the system of linear inequalities is All values that satisfy y<1/3-3 are solutions. (option 2)

For more references on inequality, click;

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I think the best answer would probably be 1

Peyton Manning threw for 1,126 yards in his first four games. At this rate, how many yards will he throw for in sixteen games?

Answers

The answer 4504 yards. You divide 1126 by 4 then multiply that answer by 16.

In how many ways can a subcommittee of 6 students be chosen from a committee which consists of 10 senior members and 12 junior members if the team must consist of 4 senior members and 2 junior members?

Answers

Answer:

The number of ways is 13860 ways

Step-by-step explanation:

Given

Senior Members = 10

Junior Members = 12

Required

Number of ways of selecting 6 students students

The question lay emphasis on the keyword selection; this implies combination

From the question, we understand that

4 students are to be selected from senior members while 2 from junior members;

The number of ways is calculated as thus;

Ways = Ways of Selecting Senior Members * Ways of Selecting Junior Members

$Ways = ^(10)C_4 * ^(12)C2$

$Ways = (10!)/((10-4)!4!)) * (12!)/((12-2)!2!))$

$Ways = (10!)/((6)!4!)) * (12!)/((10)!2!))$

$Ways = (10 * 9 * 8 * 7 *6!)/((6! * 4*3*2*1)) * (12*11*10!)/((10!*2*1))$

$Ways = (10 * 9 * 8 * 7)/(4*3*2*1) * (12*11)/(2*1)$

$Ways = (5040)/(24) * (132)/(2)$

$Ways = 210 * 66$

$Ways = 13860$

Hence, the number of ways is 13860 ways

There are 12 squares and 9 circles . What is the simplest ratio of circles to total shapes?

Answers

:) sorry i put the wrong answer before bc i thought i knew it but it was the wrong one and i dont know how to delete the answer sorrryyyyyyyyyyyyyyyy

Tom and Philip were given the graph of a linear function and asked to find the slope. Tom says that the slope is 1
2
while Philip says that the slope is 2.

Which reason correctly justifies Tom's answer?

Answers

Answer:

the answer is d

Step-by-step explanation:

i took it on usatestprep.

It’s no more to the problem?

Events A and B are mutually exclusive. Suppose event A occurs with probability 0.02 and event B occurs with probability 0.73. Compute the probability that B occurs or A does not occur (or both). Compute the probability that either B occurs without A occurring or A and B both occur

Answers

Answer:

1. The probability that B occurs or A does not occur (or both) is 0.73.

2. The probability that either B occurs without A occurring or A and B both occur is 0.73.

Step-by-step explanation:

It is given that the events A and B are mutually exclusive. It means the intersection of A and B is 0.

$P(A\cap B)=0$

Given information:

$P(A)=0.02$

$P(B)=0.73$

We get,

$P(A')=1-P(A)=1-0.02=0.98$

$P(B')=1-P(B)=1-0.73=0.27$

(1) We need to find the probability that B occurs or A does not occur (or both).

$P(B\cup A')+P(A\cap B)=P(B)+0$

$P(B\cup A')+P(A\cap B)=0.73+0$

$P(B\cup A')+P(A\cap B)=0.73$

Therefore the probability that B occurs or A does not occur (or both) is 0.73.

(2) We need to find the probability that either B occurs without A occurring or A and B both occur.

$P(B\cup A')+P(A\cap B)=P(B)+0$

$P(B\cup A')+P(A\cap B)=0.73+0$

$P(B\cup A')+P(A\cap B)=0.73$

Therefore the probability that either B occurs without A occurring or A and B both occur is 0.73.

Final answer:

For mutually exclusive events A and B, the probability that B occurs or A does not occur is approximately 1.45. The probability that either B occurs without A occurring or A and B both occur is 0.73 because A and B are mutually exclusive.

Explanation:

Events A and B are defined as mutually exclusive, which means they cannot occur at the same time. Hence, the probability that A and B both occur (referred to as P(A AND B)) is 0. In this question, for the first scenario, we need to compute the probability that B occurs or A does not occur which is denoted as P(B OR not A). Since events A and B are mutually exclusive, not A occurs with probability 1 - P(A) = 0.98. So, P(B OR not A) = P(B) + P(not A) - P(B AND not A) = 0.73 + 0.98 - (0.73 * 0.98) = 1.45 (approximately).

For the second scenario, we need to calculate the probability that either B occurs without A occurring or A and B both occur which is expressed as P((A and B) OR (B and not A)). But as we know P(A and B) = 0 for mutually exclusive events, there only remains P(B and not A). Again, as A and B are mutually exclusive, we can be sure that if B is happening, A is not, so the answer is P(B) = 0.73.

Learn more about Mutually Exclusive Events here:

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