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3 years ago

Factor the expression. d2 − 4d + 4

Mathematics

2 answers:

sesenic [268]3 years ago

7 0

Answer:

(d - 2)²

Step-by-step explanation:

d² - 4d + 4 ← is a perfect square of the form

(x ± a )² = x² ± 2ax + a²

compare coefficients of like terms

- 2a = - 4 ⇒ a = 2 and a² = 4, hence

d² - 4d + 4 = (d - 2)²

ankoles [38]3 years ago

6 0

Rewrite it in the form a^2 - 2ab + b^2, where a = d and b = 2

d^2 - 2(d)(2) + 2^2

Use the Square of Difference: (a - b)^2 = a^2 - 2ab + b^2

(d - 2)^2

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Can y’all help me on question 20?!

denis23 [38]

Answer:

yes I can help you but ???

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3 years ago

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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:

$p = 0.383$

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

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

(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 $f(3) = 0.09$

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

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

7 1/6 times 3/4 is??

Paladinen [302]

Answer:

$7\frac{1}{8}$

Step-by-step explanation:

Multiply across.

$(7\frac{1}{6})(\frac{3}{4})=7\frac{3}{24}$

Then Simplify.

$7\frac{3}{24}=7\frac{1}{8}$

8 0

3 years ago

CAN SOMEONE PLEASE HELP AND EXPLAIN THIS

lord [1]

Answer:

what are the options? I'll say it in the comments

7 0

3 years ago

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strojnjashka [21]

Answer:

Step-by-step explanation:

8/v=6/v-3

8*(v-3)=6v

8v-8*3=6v

8v-24=6v

+24 +24

8v=6v+24

-6v -6v

2v=24

v=24/2

v=12 miles /hour

v2=12-3=9 miles /hour ( your friend)

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3 years ago