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In-s [12.5K]
3 years ago
6

Help me pls with this question thank uuu

Mathematics
1 answer:
madam [21]3 years ago
3 0

Let p be 2p and q be 3q. The ratio formed by this will be

\sf \: p = 2 \\  \sf \: q = 3 \\ \sf \: p : q = 2 : 3

  • <em>Thus</em><em>,</em><em> </em><em>Option</em><em> </em><em>A</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>correct</em><em> </em><em>choice</em><em>!</em><em>!</em><em>~</em>
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Use the Distributive Property to expand 9(3 + x).
tankabanditka [31]
Distribute the 9 to each term in the parentheses with multiplication.

9(3 + x) = 27 + x

But this is usually written as x + 27
4 0
3 years ago
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Evaluate the function for x = 2 and x = 6.<br> f(x) = -(x - 2)<br> need help like asap
Kay [80]

Answer:

zero and negative four

I hope that's what you meant I don't know

Step-by-step explanation:

f(x) = -(x - 2)

=-(2-2)

=0

f(x) = -(x - 2)

=-(6-2)

=-4

5 0
3 years ago
Use distributive property to simplify 5(s-5t)
Fofino [41]
<span>5(s-5t) =5s -25t ...</span><span>distributive property

hope it helps</span>
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3 years ago
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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
Read the word problem.
san4es73 [151]

Answer:


Step-by-step explanation:

10 x 3 = $30

10 x 4 = $12

$30 + $12 = $42 in think thats the answer i havent been to school in a while

5 0
3 years ago
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