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3 years ago

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Justin weighs 15 pounds less than Greg weighs. Half of Greg’s weight is 75 pounds less than Justin’s weight. How much does each

of them weigh? A. Greg weighs 200 pounds, and Justin weighs 185 pounds. B. Greg weighs 190 pounds, and Justin weighs 175 pounds. C. Greg weighs 180 pounds, and Justin weighs 165 pounds. D. Greg weighs 170 pounds, and Justin weighs 155 pounds.

1 answer:

liubo4ka [24]3 years ago

6 0

Answer:

C. Greg weighs 180 pounds, and Justin weighs 165 pounds

Step-by-step explanation:

let G be the weight of Greg and J be the weight of Justin

G -15=J

G/2 +75=J

G =J+15 (1)

G+150=2J. (2)

If We substitute G by J+15 in equation (2) we will get:

G =J+15

J+15+150=2J

G =J+15

J+15+150=2J

G =J+15

J+165=2J

then J=165 and G=165+15=180

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2 years ago

What is the probability of getting two heads and 1 tail

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3 years ago

Read 2 more answers

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

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3 years ago

A number is equal to twice a smaller number plus 3. Yhe same number is equal to twice the sum of smaller number and 1. How many

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Answer:

There is no solution to the problem

Step-by-step explanation:

In order to find the solutions to the problem you can write the situation in an algebraic form.

You have that a number is equal to twice a smaller number plus 3. This can be written as follow:

$x=2y+3$ (1)

where x is the number and y is the smaller number

Furthermore, the same number x is equal to twice the sum of the smaller number and 1, which can be written as follow:

$x=2(y+1)$ (2)

To find the solution you equal the equation (1) with (2), and you solve for y:

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