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3 years ago

PLZZZZZZ HELPPP question:What is the mode of this data?

Mathematics

1 answer:

wariber [46]3 years ago

4 0

Answer:

The mode would be 2 because it has the higher amount of x's

I hope this is helpful

Step-by-step explanation:

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Ksenya-84 [330]

Answer:

G is the answer

Step-by-step explanation:

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2 years ago

Solve 7x − 2(x + 1) = 6x + 14.<br> −12<br> 12<br> −16<br> 16

Vladimir79 [104]

7x-2(x+1)=6x+14
7x-2x-2=6x+14
5x-6x=14+2
-x=16
x=-16

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2 years ago

Read 2 more answers

Complete this quiz to finish the assignment for 3A - Geometry Pap due on Mar 23, 2021

Natalija [7]

Answer:

What is the Picture?

Step-by-step explanation:

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3 years ago

I need help with this

Trava [24]

Answer:

https://www.mathpapa.com/algebra-calculator.html

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2 years ago

An urn contains 12 balls, of which 4 are white.

wel

Answer:

a) $\frac{1}{3}$

b) $\frac{4}{12}\,,\,\frac{4}{11}\,,\,\frac{4}{10}\,,...$

Step-by-step explanation:

Number of balls in the urn = 12

Number of white balls in the urn = 4

So, number of balls which are not white = 12 - 4 = 8

We know that probability = No. of outcomes/Total number of outcomes

Let $A_i\,,\,B_i\,,\,C_i$ denotes the event of getting white ball by A, B and C

a) each ball is replaced after being drawn.

Probability = Number of white balls/Total number of balls

Solution: $P(A_i)=P(B_i)=P(C_i)=\frac{4}{12}=\frac{1}{3}$

b)the balls that are withdrawn are not replaced.

Solution:

If A wins: $P(A_1)=\frac{4}{12}=\frac{1}{3}$

If A lost and B wins: $P\left ( B_1|A_1^{c} \right )=\frac{4}{11}$

If A and B lost and C wins: $P\left ( C_1|A_1^{c}\,B_1^{c} \right )=\frac{4}{10}$

and so on....

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3 years ago