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andrew11 [14]
3 years ago
13

Can someone use FOIL for this? pls help

Mathematics
1 answer:
Reptile [31]3 years ago
8 0

Answer:

xy^4+2x^3y^2+xy

Step-by-step explanation:

shown above. Hope this helps :]

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Help please help<br>thanks​
777dan777 [17]

Answer:

1. 1 x 10^11

2. 4.9 x 10^15

3. 1.4 x 10^4

4. 2.07 x 10^-5

5. 4 x 10^17

8 0
3 years ago
Could somebody PLZZZZ help me. Correct answers get brainly
Angelina_Jolie [31]
X(x+4)
(x+2)^2-4
(x+4)x
Hope this helps.
5 0
3 years ago
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
Someone plssssssssssssssssssss helpppppppppppppppppp
GarryVolchara [31]

Answer:

157.62

Step-by-step explanation:

7 0
3 years ago
|x^2+x-1|=1 solve for x
Marizza181 [45]

<span><span>x=<span>−3</span></span><span>x=<span>-3</span></span></span>

Here the steps

<span><span><span>2<span>x+1</span></span>−<span>1<span>x−1</span></span>=<span><span>2x</span><span><span>x2</span>−1</span></span></span><span><span>2<span>x+1</span></span>-<span>1<span>x-1</span></span>=<span><span>2x</span><span><span>x2</span>-1</span></span></span></span>

 <span>x<span>−3</span></span><span><span>(<span>x+1</span>)</span><span>(<span>x<span>−1</span></span>)</span></span>=<span><span>2x</span><span><span>(<span>x+1</span>)</span><span>(<span>x<span>−1</span></span>)</span></span></span><span><span><span>x<span>-3</span></span><span><span>x+1</span><span>x<span>-1</span></span></span></span>=<span><span>2x</span><span><span>x+1</span><span>x<span>-1</span></span></span></span></span>

<span><span><span>(<span>x+1</span>)</span><span>(<span>x<span>−1</span></span>)</span></span><span><span>x+1</span><span>x<span>-1</span></span></span></span>

<span><span><span>x<span>−3</span></span>=<span>2x</span></span><span><span>x<span>-3</span></span>=<span>2x</span></span></span>

<span><span>x=<span>−3</span></span><span>x=<span>-<span>3</span></span></span></span>

<span><span><span><span>So the answer comes out to be x = 3 I hope this was helpful </span></span></span></span>

3 0
4 years ago
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