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3 years ago

I need help please i do not understand it.

Mathematics

1 answer:

Whitepunk [10]3 years ago

7 0

Answer:

1. b=28 m

2. c=34 m

3. Yes

4. No

5. Yes

6. Yes

7. Yes

8. 6.7

9. 7.1

10. 7.1

11. The Pythagorean Theorem can be used to tell the distance between two points, which can be useful when building or traveling.

Step-by-step explanation:

The Pythagorean Theorem is just a^2+b^2=c^2

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The length of a rectangle is six times its width. If the rectangle is 600 yd^2 find its perimeter

vovangra [49]

Answer:

8400 yd^2

Step-by-step explanation:

600x 6 because its six times

3600

L+L+W+W=p

8400 yd ^2

4 0

3 years ago

What is the degree of this polynomial? 8x^6-10x+9

gtnhenbr [62]

The degree is the biggest power present in an equation.
The first is 6.
The second is 5.
Be careful with equations that aren't fully expanded.
Hope this helps

3 0

3 years ago

2x - 4y = -2<br>2x + 3y = 12

Iteru [2.4K]

Answer: x=3 and y=2

Your welcome

8 0

3 years ago

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Fr.ee points and offering brainliest for the best reply<br> and what is 1 + 1?<br> 25 points

nikitadnepr [17]

Answer:

2, your most gracious highness, superior leader of the Northern hemisphere

Step-by-step explanation:

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3 years ago

Read 2 more answers

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