DE is a midsegment of ABC . Find the value of x.

Karolyn is arranging her art project, science project, and math project on her desk. The tree diagram below shows the sample spa

murzikaleks [220]

Answer would be two over six

3 0

4 years ago

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The polygon given below is a regular pentagon.

Tpy6a [65]

Answer:

Step-by-step explanation:

In a regular pentagon, all the angles and sides are equal. So, to find an angle, divide 540 by 5.

Sum of all angles of regular pentagon = 540

measurement of one angle = 540 ÷ 5

= 108°

∠N = 108°

4 0

2 years ago

Find the tangent of angle Θ in the triangle below.

Rina8888 [55]

Answer: $\frac{\sqrt{65}}{4}$

Step-by-step explanation:

By the Pythagorean theorem, the unknown side has length

$\sqrt{9^{2}-4^{2}}=\sqrt{65}$

Therefore,

$\tan \theta=\boxed{\frac{\sqrt{65}}{4}}$

4 0

2 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

(Do not use spaces. Use to represent exponents. Example 2^3 is 22.)

Alik [6]

Answer: y=6^x-3

It is a exponent form of graph, so first:

y=a^x-b

When b=0, the asymptote is y=0 but as the asymptote given is y=-3, b=-3

Second:

the y value increases 6, when x changes 0 to 1, so a=6

8 0

3 years ago

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