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3 years ago

Jason’s mom gave him 1,000 for his birthday. He put the money in the bank and decides to withdraw 25 per week.

Mathematics

1 answer:

Goshia [24]3 years ago

6 0

Well 1000 for his birthday then subtract 25 each week, w, should be 1000-25w if i am not mistaken.

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Ierofanga [76]

Answer:

see below

Step-by-step explanation:

m/3 + 4 = 7

Subtract 4 from each side

m/3+4-4 = 7-4

m/3 = 3

Multiply each side by 3

m/3*3 = 3*3

m =9

4 0

2 years ago

Josh worked 30 hours last week and earned $8.50 per hour. If $32.50 was withheld from his paycheck, what was Josh's net pay?

MA_775_DIABLO [31]

Answer:

$222.50

Step-by-step explanation:

Hours Josh worked last week = 30

Money earned per hour = $8.50

Money withheld = $32.50

Josh's net pay = (30 x 8.50) - 32.50

= 255 - 32.50

= $222.50

Pls mark as brainliest <3

4 0

2 years ago

A TV has an original price of $549. Enter the new price after the given percent of

Free_Kalibri [48]

Answer:

384.3

Step-by-step explanation:

6 0

2 years ago

In a particular game, a fair die is tossed. If the number of spots showing is either four or five, you win $1. If the number of

TiliK225 [7]

Answer:

The probability that you win at least $1 both times is 0.25 = 25%.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either you win at least $1, or you do not. Games are independent. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of winning at least $1 on a single game:

The die has 6 sides.

If it lands on 4, 5 or 6(either of the three sides), you win at least $1. So

$p = \frac{1}{2} = 0.5$

You are going to play the game twice.

This means that $n = 2$

The probability that you win at least $1 both times is

This is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.5)^{2}.(0.5)^{2} = 0.25$

The probability that you win at least $1 both times is 0.25 = 25%.

4 0

2 years ago

Michael sold his painting for $50. He sold his painting for $10 less than its original price. At what price did Michael buy the

netineya [11]

It would be 60. He sold his painting for 50 which was 10 less that the original price. So 50+10=60

7 0

2 years ago

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