Given the replacement set {0, 1, 2, 3, 4), solve 7x-3 = 18.

What are the coordinates of A’, the image of A(-7, 9), after a rotation of -180° about the origin?

Dmitry_Shevchenko [17]

Answer:

7,-9

Step-by-step explanation:

4 0

3 years ago

NEED HELPPPPPP NUMBER 15 answer please

Harlamova29_29 [7]

It's Riverview but don't know about the percentage, I do know that it has a increase of 20k more than the rest

8 0

3 years ago

Read 2 more answers

A random sample of 4000 college students yielded 2250 who are in favor of banning Hawaiian shirts. Estimate the true proportion

Andreas93 [3]

Answer: (0.5496, 0.5754)

Step-by-step explanation:

The confidence interval for population proportion (p) is given by :-

$\hat{p}\pm z*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ , where $\hat{p}$ = Sample proportion , n= sample size , z*= Critical z-value.

Let p be the proportion of all college students who are in favor of banning Hawaiian shirt.

Given, A random sample of 4000 college students yielded 2250 who are in favor of banning Hawaiian shirts.

i.e. n=4000

$\hat{p}=\dfrac{2250}{4000}=0.5625$

z-value for 90% confidence level is 1.645

Now , 90% confidence interval for p would be :

$0.5625\pm (1.645)(\sqrt{\dfrac{0.5625(1-0.5625)}{4000}})$

$=0.5625\pm (1.645)(\sqrt{0.0000615234375})\\\\=0.5625\pm (1.645)(0.00784368774876)\\\\\approx0.5625\pm0.0129\\\\=(0.5625-0.0129, \ 0.5625+0.0129)\\\\=(0.5496,\ 0.5754)$

Hence, the required 90% interval = (0.5496, 0.5754)

6 0

4 years ago

Lila grabed the points -2 and 2 on a number line .what does the distance between these two points represent

riadik2000 [5.3K]

The distance to this question is 4

8 0

4 years ago

You play in a soccer tournament, that consists of 5 games. Each game you win with probability .6, lose with probability .3, and

nasty-shy [4]

Answer:

(a) The joint PMF of W, L and T is:

$P(W,\ L,\ T)={5\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [0.60]^{n_{W}}\times [0.30]^{n_{L}}\times [0.10]^{n_{T}}$

(b) The marginal PMF of W is:

$P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}$

Step-by-step explanation:

Let X = number of soccer games played.

The outcome of the random variable X are:

W = if a game won

L = if a game is lost

T = if there is a tie

The probability of winning a game is, P (W) = 0.60.

The probability of losing a game is, P (L) = 0.30.

The probability of a tie is, P (T) = 0.10.

The sum of the probabilities of the outcomes of X are:

P (W) + P (L) + P (T) = 0.60 + 0.30 + 0.10 = 1.00

Thus, the distribution of W, L and T is a appropriate probability distribution.

(a)

Now, the outcomes W, L and T are one experiment.

The distribution of n independent and repeated trials, each having a discrete number of outcomes, each outcome occurring with a distinct constant probability is known as a Multinomial distribution.

The outcomes of X follows a Multinomial distribution.

The joint probability mass function of W, L and T is:

$P(W,\ L,\ T)={n\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [P(W)]^{n_{W}}\times [P(L)]^{n_{L}}\times [P(T)]^{n_{T}}$

The soccer tournament consists of n = 5 games.

Then the joint PMF of W, L and T is:

$P(W,\ L,\ T)={5\choose (n_{W}!\times n_{L}!\times n_{T}!)}\times [0.60]^{n_{W}}\times [0.30]^{n_{L}}\times [0.10]^{n_{T}}$

(b)

The random variable W is defined as the number games won in the soccer tournament.

The probability of winning a game is, P (W) = p = 0.60.

Total number of games in the tournament is, n = 5.

A game is won independently of the others.

The random variable W follows a Binomial distribution.

The probability mass function of W is:

$P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}$

Thus, the marginal PMF of W is:

$P(W=w)={5\choose n_{W}!}\times 0.60^{n_{W}}\times (1-0.60)^{n-n_{W}}$

3 0

3 years ago