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2 years ago

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1. Joshua has a ladder that is 14 ft long. He wants to lean the ladder against a vertical wall so that the top of the ladder is

13.8 ft above the ground. For safety reasons, he wants the angle the ladder makes with the ground to be no greater than 78°. Will the ladder be safe at this height? Show your work and draw a diagram to support your answer.

1 answer:

34kurt2 years ago

8 0

Answer:

Step-by-step explanation:

θ is the angle the ladder makes with the ground.

θ = arcsin(13.8/14) ≈ 80.3° > 78°

The ladder is not safe at that height and angle.

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2 years ago

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3 years ago

Mark studies a group of 30 whales. If each whale weighed approximately 3.8 x 105 pounds, find the total weight of all 30 whales.

Yakvenalex [24]

Answer:

$= 1.14 \times 10^{7}pounds$

Step-by-step explanation:

If each whales weighed

$3.8 \times {10}^{5}$

Then 30 whales weighed

$30 \times 3.8 \times {10}^{5}$

$=114 \times {10}^{5}$

$= 1.14 \times 10^{2} \times {10}^{10}$

Applying the law of indices we obtain

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3 years ago

3/8 + 1/4 + 1/2 - 2/3 =

Nostrana [21]

Answer:

$\frac{11}{24}$

Step-by-step explanation:

3/8 + 1/4 + 1/2 - 2/3

- > 1/4 = 2/8

3/8 + 2/8 + 1/2 - 2/3

5/8 + 1/2 - 2/3

- > 1/2 = 4/8

5/8 + 4/8 - 2/3

9/8 - 2/3

- > LCM of 8,3: 24

- > 9/8 = 27/24

- > 2/3 = 16/24

27/24 - 16/24

11/24

Hope this helps you.

7 0

3 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

2 years ago