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4 years ago

11

Simplify: 1. Write the prime factorization of the radicand. 2. Apply the product property of square roots. Write the radicand as

a product, forming as many perfect square roots as possible. 3. Simplify. =

2 answers:

Elis [28]4 years ago

4 0

Simplify:

<span>1. Write the prime factorization of the radicand.</span> <span>2. Apply the product property of square roots. Write the radicand as a product, forming as many perfect square roots as possible. </span>

3. Simplify.

=the answer is 18

KonstantinChe [14]4 years ago

3 0

Here's the answer to something that won't let me copy and paste the question. Ur welcome.

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Please answer this correctly

ira [324]

Answer:

Surface area = 804m^2

5 0

3 years ago

Question 2 of 10

AnnyKZ [126]

The length of AB will be 10 units. Option B is corect. The formula for the distance between the two points is applied in a given problem.

<h3>What is the distance between the two points?</h3>

The length of the line segment connecting two places is the distance between them.

The distance between two places is always positive, and equal-length segments are referred to as congruent segments.

The given coordinate in the problem is;

(x₁,y₁)=(-2,-4)

(x₂, y₂)= (-8, 4)

The distance between the two points is found as;

$\rm d= \sqrt{(-8-(-2))^2+(4-(-4))^2}\\\\ d=10 \ units$

Hence, option B is corect.

To learn more about the distance between the two points, refer to;

brainly.com/question/16410393

#SPJ1

4 0

2 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago

A fruit company delivers its fruit in two types of boxes: large and small. a delivery of 5 large boxes and 3 small boxes has a t

Hoochie [10]

Let's let the weight of a large box be L, and the weight of a small box be S.

We know that 5 large boxes and 3 small boxes is 120kg, so:
5L + 3S = 120

We also know that 7 large boxes and 9 small boxes is 234kg, so:
7L + 9S = 234

You can multiply the first equation by 3 to get:
15L + 9S = 360

See how now both equations have 9S? We can now subtract one from the other:
(15L+9S) - (7L+9S) = 360-234
8L = 126
L = 15.75

Now sub this value back into an equation:
(5x15.75) + 3S = 120
3S = 41.25
S = 13.75

Double check these values
(7x15.75) + (9x13.75)
= 110.25 + 123.75
=234, which is consistent with above.

So a large box is 15.75kg, and a small box is 13.75kg.

Hope this helped

5 0

3 years ago

A random sample of 49 statistics examinations was taken. the average score, in the sample, was 84 with a variance of 12.25. the

tester [92]

Since in this case we are only using the variance of the sample and not the variance of the real population, therefore we use the t statistic. The formula for the confidence interval is:

<span>CI = X ± t * s / sqrt(n) ---> 1</span>

Where,

X = the sample mean = 84

t = the t score which is obtained in the standard distribution tables at 95% confidence level

s = sample variance = 12.25

n = number of samples = 49

From the table at 95% confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t is around:

t = 1.68

Therefore substituting the given values to equation 1:

CI = 84 ± 1.68 * 12.25 / sqrt(49)

CI = 84 ± 2.94

CI = 81.06, 86.94

<span>Therefore at 95% confidence level, the scores is from 81 to 87.</span>

6 0

4 years ago