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2 years ago

0.7 (3s + 4)-1.1s = 7.9

2 answers:

5 0

First multiply 0.7 by (3s+4) to get 2.1s + 2.8 and then add the 1.1s to the 2.1s and that should leave you with 1s+2.8 now you should have 1s + 2.8 = 7.9, you want to take 2.8 from both sides to bring it over to the 7.9 to make it look like 1s = 7.9 - 2.8 so now go ahead and have the 7.9 and the 2.8 subtract so you should end up with 1s = 5.1 so now you want to divide both sides by 1 and the 1s would cancel out into s and the 5.1 stays the same so now you have s = 5.1, you can always check this answer by seeing that it works so here is how you do that. Plug in your number, so you have
0.7 (3 * 5.1+4) - 1.1 * 5.1 = 7.9
here is your verification, 0.7 multiplied by all the numbers in the parentheses (3 * 5.1+4) would get you with 10.71 + 28 and you have - 1.1 * 5.1 = 7.9 so now just add the 10.71 to the 2.8 to get 13.51 and multiply the 1.1 by the 5.1 to get 5.61 and subtract the two to get 7.9 and that’s 7.9 = 7.9 so I hope this was helpful!! ^_^

vichka [17]2 years ago

3 0

Answer:

s = 5.1

Step-by-step explanation:

$0.7 (3s + 4)-1.1s = 7.9\\$

Multiply both sides of the equation by 10

$0.7\left(3s+4\right)\cdot \:10-1.1s\cdot \:10=7.9\cdot \:10\\\\= 7\left(3s+4\right)-11s=79\\$

Expand

$\mathrm{Expand\:}7\left(3s+4\right)-11s:\\\quad 10s+28$

Subtract 28 from both sides

$10s+28-28=79-28$

Simplify

$10s=51$

Divide both sides of the equation by 10

$\frac{10s}{10}=\frac{51}{10}$

Simplify

$s=\frac{51}{10}\\\\Decimal = 5.1$

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You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white

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Answer:

Part a: <em>The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b: <em>The case in such a way that the chances are maximized so the case where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c: <em>The minimum and maximum probabilities of winning for n number of balls are such that </em>

<em>when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n</em>

<em>when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5</em>

Step-by-step explanation:

Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as

P(A)=P(A')=0.5

Now the probability of finding the black ball is given as

P(B)=P(B∩A)+P(P(B∩A')

P(B)=(P(B|A)P(A))+(P(B|A')P(A'))

Now there can be four cases as follows

Case 1: When all the four balls are in urn A and no ball is in urn A'

P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.25*0.5)+(0*0.5)

P(B)=0.125;

Case 2: When the black ball is in urn A and 3 white balls are in urn A'

P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1*0.5)+(0*0.5)

P(B)=0.5;

Case 3: When there is 1 black ball and 1 white ball in urn A and 2 white balls are in urn A'

P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.5*0.5)+(0*0.5)

P(B)=0.25;

Case 4: When there is 1 black ball and 2 white balls in urn A and 1 white ball are in urn A'

P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.33*0.5)+(0*0.5)

P(B)=0.165;

Part a:

<em>As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b:

<em>As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c:

The minimum and maximum probabilities of winning for n number of balls are such that

when all the n balls are placed in one of the urns the probability of the winning will be least given as

P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/n*1/2)+(0*0.5)

P(B)=1/2n;

when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as

P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/1*1/2)+(0*0.5)

P(B)=0.5;

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3 years ago