Is the ordered pair (4, -2) a solution to this system? Explain why or why not.

What is the sq root of 66666647362617?

PIT_PIT [208]

Answer:

8164964.62715

Step-by-step explanation:

please mark this answer as brainliest

4 0

2 years ago

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4.Given w(x) =3×-2, v(x) =2×+7 and k(x)= -6x-7, find (w-v-k) (2)?

Ivan

Answer:

12

Step-by-step explanation:

We input 2 into each function:

w(2) = 3(2) - 2 = 4

v(2) = 2(2) + 7 = 11

k(2) = -6(2) - 7 = -19

Then we place our answers in w - v - k

(4) - (11) - (-19)

= 4 - 11 + 19

= 12

3 0

2 years ago

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The table shows the participation in a school’s debate club during the last four years.

Dominik [7]

Incomplete question, but use the following example as an illustration.

Step-by-step explanation:

Say one person had $200 in a savings account last month and now has $225. That's an increase. The problem is to find the percent of increase in money.

Read more on Brainly.com - brainly.com/question/14791979#readmore

7 0

3 years ago

Can I get the answer for 3 4 5 6

lisov135 [29]

3) x=190, <BOC=85

4) x=177, <TOU=31

5) x=61, <LOM=110

6) x=55, <DOE=117

4 0

3 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

3 years ago