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3 years ago

11

An angle that is greater then 90 degrees and less then 180degrees

1 answer:

salantis [7]3 years ago

4 0

It would be an obtuse angle :)

You might be interested in

What is an equation of the line that passes through the points (-5,-1) and (5,3)?

Mademuasel [1]

Answer:

The equation of the line is:

$y=\frac{2}{5}x+1$

Step-by-step explanation:

Given the points

(-5, -1)

(5, 3)

Finding the slope between points

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-5,\:-1\right),\:\left(x_2,\:y_2\right)=\left(5,\:3\right)$

$m=\frac{3-\left(-1\right)}{5-\left(-5\right)}$

$m=\frac{2}{5}$

Using the point-slope form to find the line equation

$y-y_1=m\left(x-x_1\right)$

substituting the values m = 2/5 and the point (5, 3)

$y-3=\frac{2}{5}\left(x-5\right)$

Add 3 to both sides

$y-3+3=\frac{2}{5}\left(x-5\right)+3$

$y=\frac{2}{5}x+1$

Thus, the equation of the line is:

$y=\frac{2}{5}x+1$

4 0

3 years ago

Please help me out with this

gavmur [86]

All of them?
Or just the ones in the mddle

4 0

3 years ago

25% of Rebekah's movies are Marvel. If she has 85 Marvel movies, how many movies does she own?

AleksandrR [38]

Answer:

340

Step-by-step explanation:

25% is also 25/100 so you just need to times 85x4 to get the answer. The answer to that is 340.

i think this is right.

3 0

3 years ago

HELP ASAP!!! I'M TIMED This year will have a February 29th if, and only if, this is a leap year.

earnstyle [38]

If and only if is the phrase that makes this biconditional

7 0

3 years ago

Read 2 more answers

A total of 2n cards, of which 2 are aces, are to be randomly divided among two players, with each player receiving n cards. Each

klasskru [66]

Answer:

$P(X_s^c|X_F) =0.2$

$P(X_s^c|X_F) =0.31$

$P(X_s^c|X_F) =0.331$

Step-by-step explanation:

From the given information:

Let represent $X_F$ as the first player getting an ace

Let $X_S$ to be the second player getting an ace and

$\sim X_S$ as the second player not getting an ace.

So;

The probabiility of the second player not getting an ace and the first player getting an ace can be computed as;

$P(\sim X_S| X_F) = 1 - P(X_S|X_F)$

$P(X_S|X_F) = \dfrac{P(X_SX_F)}{P(X_F)}$

Let's determine the probability of getting an ace in the first player

i.e

$P(X_F) = 1 - P(X_F^c)$

$= 1 -\dfrac{(^{2n-2}_n)}{(^{2n}_n)}}$

$= 1 - \dfrac{n-1}{2(2n-1)}$

$= \dfrac{3n-1}{4n-2} --- (1)$

To determine the probability of the second player getting an ace and the first player getting an ace.

$P(X_sX_F) = \text{ (distribute aces to both ) and (select the left over n-1 cards from 2n-2 cards}$ $P(X_sX_F) = \dfrac{2(^{2n-2}C_{n-1})}{^{2n}C_n}$

$P(X_sX_F) = \dfrac{n}{2n -1}---(2)$

$P(X_s|X_F) = \dfrac{2}{1}$

$P(X_s|X_F) = \dfrac{2n}{3n -1}$

Thus, the conditional probability that the second player has no aces, provided that the first player declares affirmative is:

$P(X_s^c|X_F) = 1- \dfrac{2n}{3n -1}$

$P(X_s^c|X_F) = \dfrac{n-1}{3n -1}$

Therefore;

for n= 2

$P(X_s^c|X_F) = \dfrac{2-1}{3(2) -1}$

$P(X_s^c|X_F) = \dfrac{1}{6 -1}$

$P(X_s^c|X_F) = \dfrac{1}{5}$

$P(X_s^c|X_F) =0.2$

for n= 10

$P(X_s^c|X_F) = \dfrac{10-1}{3(10) -1}$

$P(X_s^c|X_F) = \dfrac{9}{30 -1}$

$P(X_s^c|X_F) = \dfrac{9}{29}$

$P(X_s^c|X_F) =0.31$

for n = 100

$P(X_s^c|X_F) = \dfrac{100-1}{3(100) -1}$

$P(X_s^c|X_F) = \dfrac{99}{300 -1}$

$P(X_s^c|X_F) = \dfrac{99}{299}$

$P(X_s^c|X_F) =0.331$

8 0

3 years ago