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3 years ago

The sum of 8 times a number and 3 equals 5

Mathematics

2 answers:

Burka [1]3 years ago

3 0

8n+3=5
8n=2
N=.25

The number is .25 or 1/4

Lemur [1.5K]3 years ago

3 0

Answer:

im not sure if it can be a decimal tell me if it cant but if it can its 0.25

Step-by-step explanation:

0.25*8=2 2+3=5

and i get my answer by a little bit of trial and error. I knew that what ever i timed by 8 had to equal 2 in order for the +3 to get it to 5 and the only thing that you can times by 8 and get 2 is 0.25.

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Find the vertical asymptote(s) of f of x equals quantity 4 x squared plus 3x plus 6 end quantity over quantity x squared minus 3

siniylev [52]

The vertical asympototes of f(x) are at x = -6 and x = 6

Step-by-step explanation:

To find the vertical asymptote(s) of a rational function,

Equate the denominator by 0
Solve it for x
If x = a, then the vertical asymptote is at x = a

∵ $f(x)=\frac{4x^{2}+3x+6}{x^{2}-36}$

- Equate the denominator x² - 36 by 0

∵ x² - 36 = 0

- Add 36 to both sides

∴ x² = 36

- Take √ for both sides

∴ x = ± 6

∴ There are vertical asymptotes at x = -6 and x = 6

The vertical asympototes of f(x) are at x = -6 and x = 6

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3 years ago

In the picture below, what is the value of x?

lana [24]

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5 0

3 years ago

Evaluate 9ab, for a = 5 and b =<br> - 7

Yuri [45]

Answer:

-315

Step-by-step explanation:

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7 0

3 years ago

Read 2 more answers

Let X be a Bernoulli rv with pmf as in Example 3.18. a. Compute E(X2 ). b. Show that V(X) 5 p(1 2 p). c. Compute E(X79).

spayn [35]

The Bernoulli distribution is a distribution whose random variable can only take 0 or 1

The value of E(x2) is p
The value of V(x) is p(1 - p)
The value of E(x79) is p

<h3>How to compute E(x2)</h3>

The distribution is given as:

p(0) = 1 - p

p(1) = p

The expected value of x2, E(x2) is calculated as:

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * (1- p) + 1^2 * p$

Evaluate the exponents

$E(x^2) = 0 * (1- p) + 1 * p$

Multiply

$E(x^2) = 0 +p$

Add

$E(x^2) = p$

Hence, the value of E(x2) is p

<h3>How to compute V(x)</h3>

This is calculated as:

$V(x) = E(x^2) - (E(x))^2$

Start by calculating E(x) using:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * (1- p) + 1 * p$

$E(x) = p$

Recall that:

$V(x) = E(x^2) - (E(x))^2$

So, we have:

$V(x) = p - p^2$

Factor out p

$V(x) = p(1 - p)$

Hence, the value of V(x) is p(1 - p)

<h3>How to compute E(x79)</h3>

The expected value of x79, E(x79) is calculated as:

$E(x^{79}) = \sum x^{79} * P(x)$

So, we have:

$E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p$

Evaluate the exponents

$E(x^{79}) = 0 * (1- p) + 1 * p$

Multiply

$E(x^{79}) = 0 + p$

Add

$E(x^{79}) = p$

Hence, the value of E(x79) is p

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Ava has been asked to create the advertising art for the school skating party. She is using paper that is 15.5 inches wide. The

marshall27 [118]

15.5in -0.75in= 14.75in No, because the left over space on the paper with the border is 14.75in wide and the design is 14.5in wide.

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