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OLEGan [10]
4 years ago
6

Rewrite the equation of a line in point-slope form to slope-intercept form: Y-3=1/2(x-8)

Mathematics
2 answers:
zzz [600]4 years ago
6 0
Y=1/2 (8-x)+3 if you're doing y=mx+b
Drupady [299]4 years ago
5 0
(1/2)x - 1 should be the answer
Distribute 1/2 to x and 8 which gives you (1/2)x and - 4
(1/2)x - 4 = Y - 3
Bring - 3 to the other side
(1/2)x - 4 +3
(1/2)x - 1 = Y
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It is known that 4 <br> PLS HELP
sergejj [24]

Answer:

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Step-by-step explanation:

4 0
4 years ago
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What 175 multiplyed by 260
jolli1 [7]

Answer:

45500

Step-by-step explanation:

175 can be multiplied by 200 which gives you 35000 plus 175 times 60 which gives you 10500. You can add and you get 45500.

I was a first place champion in number sense and thats my strategy

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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
4 years ago
Helpp<br> multiply with monomial pt 2
olya-2409 [2.1K]

Answer:

The answer is C

Step-by-step explanation:

Multiply it :

(-4m³z⁴)(2m²z²)

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6 0
3 years ago
Help me please I’m not good at math
Deffense [45]

Answer:

i think the answer is 7/4

6 0
3 years ago
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