Ask question

Ask question

All categories

3 years ago

13

Kate is making dresses she use 1/3 yard if fabric for each dress she makes, what is the total of fabric Kate need to make 96 dre

sses Fast I Need HELP!!!

2 answers:

algol133 years ago

6 0

It’s 32.. bc you take 96 and times it by 1/3 yards

Fantom [35]3 years ago

4 0

32 yards

96/3 is 32 and since you have a foot per dress that’s the work

You might be interested in

Which polygon in the diagram cannot be mapped onto the others by similarity transformations?

Rom4ik [11]

Answer: B.) PQRST

Step-by-step explanation:

There are no options but if its the same question I just had on a test then this is the answer..

Hope this helps..

6 0

3 years ago

Which equation is equal up to one-hundredth of 36? Is it 0.3*0.12 or 0.36÷0.12 or 0.3*1.2

Mumz [18]

Its 360 beacuse u multiply all of those numbers

3 0

3 years ago

Read 2 more answers

Jonas’s vacuum cleaner uses 5/8 kilowatts of electricity per hour. His clothes dryer uses 4 11/16 of electricity per hour. How m

alekssr [168]

Hope this helps you out.

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

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

If g(r) =2(r-1), Find g(10)

garri49 [273]

Answer:

g(r) = 18

Step-by-step explanation:

Substitute 10 into r

g(10) = 2( (10) -1)

= 2(9)

= 18

7 0

3 years ago