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3 years ago

These three please, thank you!

Mathematics

2 answers:

Lerok [7]3 years ago

5 0

1. 0.01

2. 29.44

3. (69+6)/9

Inessa05 [86]3 years ago

3 0

The first one is 9/25
The second one is 29.44
The last one is 8

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Pls helpppppp i begggg

Artist 52 [7]

Answer

x Is less than 4

Step-by-step explanation:

It will be an open circle and the arrow is going the left direction on the number 4

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2 years ago

Suppose corn chips cost 28.5 cents per ounce. If a bag costs $3.75, how many ounces are in the bag of chips? Round to the neares

ki77a [65]

13.16oz are in the bag because $3.75÷$0.285 is 13.16oz

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3 years ago

Read 2 more answers

BRAINLIEST ASAP for BRAINLIEST!!! <br> EASY MATH! Do with explanation

natulia [17]

Step-by-step explanation:

you plug 0 into the (x,y) into the equation. if your finding x you plug in 0 for y and versa.

so y = 9x -3 im going to find x

0 = 9x -3 subtract -3 from both sides

-3 = 9x divide by 9

x = -3

4 0

3 years ago

Please help me, no links or random answers, please. When you are writing the answer please number it so I know which answer is f

meriva

Answer

A: (1,7.5)

b:(0,6)

c: because the graph is linear graph and have a constant rate of change/ slope which is -1.5

Step-by-step explanation:

5 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

4 years ago