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Mrrafil [7]
3 years ago
11

Solve for x 25x+5 10x

Mathematics
1 answer:
alekssr [168]3 years ago
4 0

Answer:

x = -1/3

Step-by-step explanation:

I think you mean 25x+5=10x

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katrin [286]

Answer:

0.75 + 0.25 = t

3 0
3 years ago
Please help equation below
qaws [65]

Answer:

3y √21 + 2 y√15

Step-by-step explanation:

To simply, we open up the bracket

We have this as;

√(189y^2) + √60y

= 3y √21 + 2 y√15

7 0
2 years ago
Suppose that you are given a bag containing n unbiased coins. You are told that n-1 of these coins are normal, with heads on one
gladu [14]

Answer:

The (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

Step-by-step explanation:

Given

Total unbiased coin = n

Normal coins =n - 1

Fake = 1

The (conditional) probability that the coin you chose is the fake coin is represented by

P(Fake | Head)

And it's calculated as follows;

P(Fake | Head) = P(Fake, Head) ÷ P(Head) ----- (1)

Where P(Fake, Head) = P(Fake) * P(Head | Fake)

P(Fake) = 1/n --- because only one is fake

P(Head | Fake) = n/n because all coins (including the fake) have head

So, P(Fake, Head) = P(Fake) * P(Head | Fake) becomes

P(Fake, Head) = 1/n * n/n

P(Fake, Head) = 1/n

P(Head) is calculated by

P(Fake) * P(Head | Fake) + P(Normal) * P(Head | Normal)

P(Fake) * P(Head | Fake) = P(Fake, Head) = 1/n (as calculated above)

P(Normal) * P(Head | Normal) = ½ * (n - 1)/n ----- considering that the coin also has a tail with equal probability as that of the head.

Going back to (1)

P(Fake | Head) = P(Fake, Head) ÷ P(Head) becomes

P(Fake | Head) = (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ (1/n + (n - 1)/2n)

= (1/n) ÷ (2 + n - 1)/(2n)

= (1/n) ÷ (1 + n)/(2n)

= (1/n) * (2n)/(1 + n)

= 2/(1 + n)

Hence, the (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

5 0
3 years ago
Write the equation of the line that passes through the points (-9, -3) and (-7,7).
tiny-mole [99]

Answer:

y=5x+42

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(7-(-3))/(-7-(-9))

m=(7+3)/(-7+9)

m=10/2

m=5

y-y1=m(x-x1)

y-(-3)=5(x-(-9))

y+3=5(x+9)

y=5x+45-3

y=5x+42

6 0
3 years ago
Charlene is a video game designer and wants to make sure that her games can be played on all types of screens. In order for the
Nikitich [7]

Answer:

The playable area has a width of 9 inches and a height of 6 inches.

Step-by-step explanation:

There are a number of ways you can get there.

1. Check the answers to see which have the right area and aspect ratio.

A: area = 24 in² — does not match 54 in²

B: area = 54 in², aspect ratio 6:9 = 2:3 — does not match 3:2 aspect ratio

C: area = 54 in², aspect ratio 9:6 = 3:2 — matches problem statement

D: area = 121.5 in² — does not match 54 in²

2. If the screen were 3:2 (inches), its area would be 6 in². The area of 54 in² is 9 times that value, so the actual screen dimensions are √9 = 3 times 3:2. That is, they are width:height = 9:6 inches — matches selection C.

3. You can write equations for width and height and solve.

w/h = 3/2

wh = 54

Substituting w=3/2·h into the second equation gives

... (3/2)h·h = 54

... h² = 36 . . . . . multiply by 2/3

... h = √36 = 6 . . . . square root, result in inches

4 0
2 years ago
Read 2 more answers
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