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3 years ago

1/2(b-3)=5/2 solve equation plzzz

Mathematics

1 answer:

Igoryamba3 years ago

5 0

Answer:

b=8

i hope that's right

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Evaluate 3^2 (6-2) times 4-6/3

dedylja [7]

3^2(6-2)*(4-6)/3
=(9*4)*(-2/3)
=36*(-2)/3
= -72/3
= -24 (answer)

4 0

3 years ago

A pair of shoes coats $25 to make. This means that you need to charge a price of atleast __ just to cover you Options : A:$25 B:

Anna71 [15]

Answer:

Step-by-step explanation:

Because they cost $25 to make so in order to cover the cost they would need to charge at least 25, but out of those choices if they wanted to make a profit they'd need to charge the $50

8 0

3 years ago

- 8y + 9x = -5<br> 8y + 7x = -75

IRINA_888 [86]

Answer:

x = -5 , y = -5

Step-by-step explanation:

- 8y + 9x = -5

8y + 7x = -75

__________

0 + 16x = - 80

x = -80 / 16

x = -5

- 8y + 9x = -5

-8y + 9(-5) = -5

-8y - 45 = -5

-8y = -5 + 45

-8y = 40

y = 40/-8

y = -5

hope it helps!

8 0

3 years ago

54 meters in 2.5 hours

Orlov [11]

21.6 meters per hour
Solve by dividing 54 by 2.5

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