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3 years ago

The sum of a number times 3 and 15. I don't understand it.

1 answer:

8 0

45.3 would be your anwser multiply 3 and to 1 and then 5 and then 1 add a zero or any numbers that are higher than 9 then put a zero and do the same but the oposite then add it and you will get 45.3 your welcome.

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Two squares have sides 12cm and 4cm respectively. What is the ratio of their sides

Nookie1986 [14]

Answer:

3 : 1

Step-by-step explanation:

We have to compare the value of two sides.

Ratio of sides = 12 : 4 = $\dfrac{12}{4}=\dfrac{4*3}{4*1}=\dfrac{3}{1}$

= 3 : 1

6 0

2 years ago

I need help with math

Iteru [2.4K]

area = pi times radius squared

area = pi times 5 squared

area = pi times 25

area = 78.54

answer : 78.54 square feet

4 0

1 year ago

is searching online for airline tickets. Two weeks ago, the cost to fly from to was 210 $. Now the cost is 305$. What is the

harkovskaia [24]

Answer:

Step-by-step explanation:

20]x]

3 0

3 years ago

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 $\binom{r}{r-1}$