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3 years ago

What is the solution set of 4x^2 - 36 = 0?

2 answers:

6 0

4x^2 - 36 = 0 ( Original equation )

4x^2 = 36 ( Add 36 to both sides )

x^2 = 9 ( Divide by 4 on both sides )

x = 3 , -3 ( Take the square root of 9 )

Solution set: { +3 , -3 } answer is x = +3 , x = -3

gizmo_the_mogwai [7]3 years ago

4 0

Add 36 to both sides and divide by 4
and take its square root the answer is = 3

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Can u explain how u answer this question?

Leto [7]

Answer:

Value of given expression $\frac{2p}{q}$ when p = -4 and q = 2 is -4.

Step-by-step explanation:

Given expression , $\frac{2p}{q}$

We have to evaluate the value of the expression when p = -4 and q = 2

Consider the expression given $\frac{2p}{q}$

To find the value a the given value just put the values in the given expression

$\frac{2p}{q}$ becomes $\frac{2(-4)}{2}$

Solve further to get its value,

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Thus, value of given expression $\frac{2p}{q}$ when p = -4 and q = 2 is -4.

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4 years ago

PLEASE I WILL MARK BRAINLIEST 22. Find the measure of angle HJK.

Irina18 [472]

Answer:

i think its 38 but i could be wrong sorry

Step-by-step explanation:

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3 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7

Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable X ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable X is normally distributed with mean, μ = 2.7 and standard deviation, σ = 0.75.

(a)

A random sample of n = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

$P(\bar X\leq 3.00)=P(\bar X$

$=P(\bar X$

$=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$=P(Z$

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

$P(2.70$

$=P(0$

*Use a z-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

$P(\bar X\leq 3.00)\geq 0.99$

$P(\bar X$

$P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$P(Z$

The value of z for the probability above is

z ≥ 2.33

Compute the value of n as follows:

$\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33$

$\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}$

$\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77$

Thus, the sample size must be at least 77.

8 0

3 years ago

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Irina18 [472]

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3 years ago

Select the proper placement for parentheses to speed up the addition for the expression 4 + 6 + 5

nata0808 [166]

Im guessing that by making 10 it will speed up the addition process so it should be
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6 0

4 years ago

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