All categories

3 years ago

-Simplify the expression. 2(-y + 9) + 8y

Mathematics

1 answer:

Nata [24]3 years ago

4 0

Answer:

first we open the brackets using two so our new expression will be -2y+ 18 + 8y collect the like terms -2 y + 8 y + 18 so your final answer will be 6 y + 18 if you have any questions regarding this just ask

You might be interested in

Angela invested $4000 in an account that earns 5% interest yearly (simple interest). She forgets about it for 12 years. What is

harina [27]

Answer:

$6400

Step-by-step explanation:

4000x.05=200× 12=2400+4000=$6400

4 0

3 years ago

If repetition of digits is allowed, how many different ways 5-digit codes are possible with the condition that the first digit s

Nesterboy [21]

Answer:

b) 27,000

Step-by-step explanation:

3 × 10 × 10 × 10 × 9 = 27000

'3' because the first digit can only be 3,6, or 9

'9' because the last digit cannot be zero, leaving only 9 options

7 0

3 years ago

How can I express logbx-(logby+logbz) as a single log?

GuDViN [60]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)\\\\
-----------------------------\\\\
log_b(x)-[log_b(y)+log_b(z)]\implies log_b(x)-[log_b(yz)]
\\\\\\
log_b\left( \cfrac{x}{yz} \right)$

3 0

4 years ago

Find the 50th term of this arithmetic sequence: 6, 13, 20, 27, . . .

sladkih [1.3K]

Answer: 349

Step-by-step explanation:

Using the arithmetic sequence, it would add up to be 349

7 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