All categories

3 years ago

PLEASE HELP ASAP!!!

Mathematics

2 answers:

wolverine [178]3 years ago

7 0

The correct answer is A

Softa [21]3 years ago

5 0

The correct answer is A because if you add 200 and 85 together to get the amount of money he has to have, and then divide that by 45, you get 13 weeks

You might be interested in

What is the expression is equivalent(23×2)×5

timurjin [86]

Answer:

230

Step-by-step explanation:

8 0

2 years ago

3. The student council is selling roses to collect funds for a charity. The graph be

dexar [7]

Answer:

$Unit = 3$

Step-by-step explanation:

Given

See attachment for graph

Required

The unit cost

Pick any point on the dots of the graph; we have:

$(x,y) = (2,6)$

The unit cost is then calculated as:

$Unit = \frac{6}{2}$

$Unit = 3$

5 0

3 years ago

The equation of a parabola is y=1/8x2−4x+41 .

Dafna1 [17]

Y=7............................................................................................................................................................

3 0

2 years ago

Read 2 more answers

What is 35 mL equals

AlexFokin [52]

Equal to what ?? what are you trying to convert it to, please type the whole question so i can better answer your question.

8 0

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 $P(E)+ P(E_{i})$

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

7 0

3 years ago