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3 years ago

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Mathematics

1 answer:

Pavlova-9 [17]3 years ago

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Solve the following equation for x<br><br> 5(x - 1) = -45

abruzzese [7]

Answer:

solution: x = -8

Step by step:

1) 5x-5+5 = -45+5 : Add 5 to both sides

2) 5x= -40 : Simplify

3) 5x/5 = -40/5 : Divide both sides

Hope that helps

5 0

2 years ago

A soccer team scores 12 points in 5 matches. They scored the same number of points in each of their first 3 matches and one more

NeX [460]

Answer:

2 points the first 3 matches then 3 points the last 2 matches

Step-by-step explanation:

12 points in 5 matches and the first 3 are all the same then the other 2 increased by 1 point

They got 2 points the first 3 matches then 3 points the last 2 matches

2 + 2 + 2 (first 3 matches) = 6

3 + 3 (last 2 matches) = 6

6 + 6 = 12

7 0

3 years ago

Which equation represents a line that passes through (2,-1/2<br>and has a slope of 3?

erma4kov [3.2K]

Answer:

y + 0.5 = 3(x - 2)

y + 0.5 = 3x - 6

y = 3x - 6.5

Step-by-step explanation:

8 0

3 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

3 years ago

Find f(x)=x^2-x^3,find f(-10)

Elanso [62]

Answer:

Value of f(-10) is 1100

Step-by-step explanation:

Given that f(x) = x² - x³

Here we need to find f(-10).

f(-10) = (-10)² - (-10)³

f(-10) = 100 - (-1000)

f(-10) = 100 + 1000

f(-10) = 1100

Value of f(-10) is 1100

3 0

2 years ago