Given the literal equation 2003-05-04-00-00_files/i0290000.jpg, with y = 4x + 2, what is the value of z if x = 1?

Please I need this before 12pm

luda_lava [24]

Answer:

The required formula is:

$\ a_{n}=a_{1}+(n-1)d$

Step-by-step explanation:

The total number of squares of the the first term = 4

The total number of squares of the the second term = 7

The total number of squares of the the third term = 10

so,

$a_1=4$

$a_2=7$

$a_3=10$

Finding the common difference d

$d=a_3-a_2=10-7=3$

$d=a_2-a_1=7-4=3$

As the common difference 'd' is same, it means the sequence is in arithmetic.

So

If the initial term of an arithmetic progression is $a_{1}$ and the common difference of successive members is d, then the nth term of the sequence $(a_n)$ is given by:

$\ a_{n}=a_{1}+(n-1)d$

Therefore, the required formula is:

$\ a_{n}=a_{1}+(n-1)d$

3 0

3 years ago

Factor the following expressions completely. Show and check all work on your own paper.

Katyanochek1 [597]

5x2 would be 10 and 10-50x would be negative 40x and 40x + 125 would be 165x

5 0

2 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

Is -8 = 8

madam [21]

No solution

The answer is D

5 0

2 years ago

Please explain. 100 / (-5) =

svp [43]

That would be -20. a negative divided by a negative number is positive, positive divided by negative is negative bc even numbers of negatives make it positive. and ofc 20 x 5 = 100, hence the ‘20’

5 0

3 years ago

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