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Ivan
3 years ago
6

Could someone help me with this question? i don't know how to really work this out Order Of Operations

Mathematics
1 answer:
lozanna [386]3 years ago
3 0

Answer:

10

Step-by-step explanation:

2[45/(11-8)^2]

=2[45/(3)^2]

=2[45/9]

=2[5]

=10

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84.8 sq in

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The area of the top and bottom are each 1/2(3.7 x 4) = 7.4, so they total 14.8

2 of the sides are 5 x 5 = 25, so they total 50

1 side is 5 x 4 = 20

14.8 + 50 + 20 = 84.8 sq in

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Simplify 3 and 16/24
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Answers should be exact decimals (please do not leave as fractions or unfinished calculations). A six-sided die (cube) with face
Ghella [55]

Answer:

(a) P = 0.0001

(b) P = 0.6561

(c) P = 0.2916

(d) P = 0.3439

(e) P = 0.2

Step-by-step explanation:

This is a probability problem.

The dice is rolled 4 times (n=4) and we calculate the probability of different outcomes.

The probability of a 6 in a roll is 0.5.

The probability of a 1, 2, 3, 4 or 5 in a roll is 0.5/5=0.1.

<u />

<u>a) Outcome: all the rolls are 2.</u>

The probability of having a 2 in a roll is 0.1, so we can calculate the probability of having a 2 in four consecutive rolls as

P(x1=2;x2=2;x3=2;x4=2)=P(x=2)^4=0.1^4=0.0001

<u>b) Outcome: none of the rolls is a 2.</u>

The probability of having any other number but 2 in 4 rolls is:

P(x1\neq2;x2\neq2;x3\neq2;x4\neq2)=P(x\neq2)^4=(1-0.1)^4=0.6561

<u>c) Outcome: exactly one roll is a 2</u>

This is the sum of the probability of having a 2 in the first, second, third or fouth roll, and others numbers in the rest of the rolls. These 4 combinations have the same probability, so we will multiply it by 4.

P(exactly \,one\,2)=4*P(x1=2;x2\neq2;x3\neq2;x4\neq2)\\\\P(exactly \,one\,2)=4*0.1*0.9*0.9*0.9=0.2916

<u>d) Outcome: at least one of the rolls is a 2</u>

In this case, is the probability of having at least one 2, is the sum of the probability of getting a 2 in the first roll, the probability of getting a 2 in the second roll, the probability of getting a 2 in the third roll and the probability of getting a 2 in the four roll:

P(x1=2)+P(x2=2)+P(x3=2)+P(x4=2)=0.1+0.9*0.1+0.9*0.9*0.1+0.9*0.9*0.9*0.1\\\\P(x1=2)+P(x2=2)+P(x3=2)+P(x4=2)=0.1+0.09+0.081+0.0729\\\\P(x1=2)+P(x2=2)+P(x3=2)+P(x4=2)=0.3439

<u>e) Outcome: either the first roll or the last roll is a 2</u>

The probability of getting a 2 in the first roll is equal to having it in a fourth roll, and its the probability of getting a 2 in a roll (multiplied by 2, beacuse there can be either in the first or in the last roll).

P(x1=2)+P(x4=2)=2*P(x=2)=2*0.1=0.2

6 0
3 years ago
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