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2 years ago

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In a paint factory, an old conveyer line has filled 50 barrels of paint, and is filling more at a rate of 2 barrels per minute.

A worker just switched on a newer line that can fill 12 barrels per minute. In a little while, the two lines will have filled an equal number of barrels. How many barrels will each line have filled?

1 answer:

bezimeni [28]2 years ago

4 0

Answer:

60 barrels

Step-by-step explanation:

y = # of barrels; x = minutes

worker: y = 12x

old line: y = 2x + 50

12x = 2x + 50

10x = 50

x = 5

y = 12(5) = 60

y = 2(5) + 50 = 60

60 barrels

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Help please help help please please please

AnnZ [28]

Answer:

C. $\frac{5}{4}$

D. $\frac{6}{7}$

Step-by-step explanation:

C. $2 - \frac{12}{16}$

$= 2 - \frac{3}{4}$

$= \frac{8 - 3}{4}$

$= \frac{5}{4}$

D. $\frac{20}{22} \times \frac{33}{35}$

$= \frac{10}{11} \times \frac{33}{35}$

$= 10 \times \frac{3}{35}$

$= 2 \times \frac{3}{7}$

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

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

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3 years ago

4 divided by what = 9

vfiekz [6]

It 36 hope i helped!!

Assume answer as x;

x÷4=9 or x4=9

Multiply 9 with 4;

x=9×4

x=36

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