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2 years ago

Find the LCD for the list of rational expressions. 2 7x +4 x2 + 5x - 14' x2 + 3x – 28

Mathematics

1 answer:

andre [41]2 years ago

7 0

the equation is made and the included values are indicated.

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Find the sum and express it in simplest form. <img src="https://tex.z-dn.net/?f=%28b%20-%203x%29%20%2B%20%28%20-%205b%20%2B%

nevsk [136]

(b - 3x) + (-5b + 6x - 5)
= b - 3x - 5b + 6x - 5
= b - 5b - 3x + 6x - 5
= -4b + 3x - 5

8 0

2 years ago

How many times greater was the temperature in Miami (45) than the lowest in San Diego (25)

Greeley [361]

Answer:

The temperature in Miami is 9/5 times the temperature in San Diego.

Step-by-step explanation:

Ratios

To compare the temperature in Miami (45) with the temperature in San Diego (25), we use the division or ratio between both numbers:

$\displaystyle \frac{45}{25}$

Simplify the fraction dividing by 5:

$\displaystyle \frac{9}{5}$

Since the relation between them is 9/5, it means the temperature in Miami is 9/5 times the temperature in San Diego.

3 0

3 years ago

Please, help and please answer my questions Lately, no-one's answering my questions-

evablogger [386]

Answer:

I think the last you already have is right. I wasn’t very good at this but I try

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Answer choices: 16/3 40/9 4/9 4/3

son4ous [18]

Answer:

0.44444444 or 4/9

Step-by-step explanation:

Since we know the value of $x$ , we can rewrite the equation replacing the variables.

$\frac{2}{3}^{2} + 3*\frac{2}{3} - 2$

Now solve. While solving, remember the order of operations.

$\frac{2}{3}^{2} + 3*\frac{2}{3} - 2 = 0.44444444$ or $\frac{4}{9}$

3 0

2 years ago

The n candidates for a job have been ranked 1, 2, 3,…, n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p

Anon25 [30]

Answer:

A. E(x) = 1/n×n(n+1)/2

B. E(x²) = 1/n

Step-by-step explanation:

The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as

P(x) = {1/n, x = 1,2...n}

Therefore,

Expectation of X

E(x) = summation {xP(×)}

= summation {X×1/n}

= 1/n summation{x}

= 1/n×n(n+1)/2

= n+1/2

Thus, E(x) = 1/n×n(n+1)/2

Value of E(x²)

E(x²) = summation {x²P(×)}

= summation{x²×1/n}

= 1/n

3 0

2 years ago