PLEASE HELP!!!!!

Mathematics

2 answers:

Debora [2.8K]3 years ago

8 0

The correct answer is; "△DEF is not a right triangle because no two sides are perpendicular." - I took the test.

Alenkasestr [34]3 years ago

7 0

<span>â–łDEF is not a right triangle because no two sides are perpendicular. There are two ways to determine if DEF is a right triangle. You can use the pythagorean theorem to see of a^2 + b^2 = c^2, or you can calculate the slope of each pair of lines to see if their product is -1. I'll demonstrate both methods, but given the available options, I suspect that the product of the slopes is the one you need. Slope method. Slope of DE = (1-(-1))/(-4-3) = 2/-7 = -2/7 Slope of EF = (-1-(-4))/(3-(-1)) = 3/4 Slope of DF = (1-(-4))/(-4-(-1)) = 5/-3 = -5/3 Since we're looking for -1 as a product and we have 2 negative slopes and 1 positive slope, we only need to check DE against EF and EF against DF. So -2/7 * 3/4 = -6/28 = -3/14. This isn't -1, so not perpendicular. 3/4 * -5/3 = -15/12 = -5/4. Still not -1, so not perpendicular. Since no sides of perpendicular to any other sides, DEF is not a right triangle. Now for the pythagorean theorem method. First, let's take a look at the length of the 3 sides of DEF. Actually, we just need to look at the square of the lengths of each side. So square of length DE = (-4 - 3)^2 + (1 - (-1))^2 = -7^2 + 2^2 = 49 + 4 = 53. square of length EF = (3-(-1))^2 + (-1-(-4))^2 = 4^2 + 3^2 = 16+9 = 25 square of length DF = (-4-(-1))^2 + (1-(-4))^2 = -3^2 + 5^2 = 9 + 25 = 34 The longest side is DE, so lets add up the squares of the other 2 sides to see if they equal the square of DE. 25 + 34 = 59. And 59 is not equal to 53. So we know that triangle DEF is NOT a right triangle. Now let's look at the options and see what's correct. â–łDEF is not a right triangle because no two sides are perpendicular. * This is a true statement. So it's the correct answer. The remaining 3 other options all claim that one side is perpendicular to another and as such are incorrect.</span>

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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 $\binom{r}{r-1}$