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2 years ago

Not sure how to do this giving away the rest of my points for it!

Mathematics

1 answer:

marysya [2.9K]2 years ago

7 0

Answer:

i THINK the Doman is 36 and the other one is 14

Step-by-step explanation:

sssssssssuper sorry if this is wrong!!!!!!!!!!!!!

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nasty-shy [4]

Answer:

The answer of this question is 50

7 0

3 years ago

Read 2 more answers

Can someone help me please

rusak2 [61]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

The sum is formed by adding corresponding elements. For example, the lower right element (row 3, col 3) of the sum is 5+4 = 9

3 0

3 years ago

WHOEVER ANSWERS FIRST GETS BRAINLIEST AND I NEED AN EXPLANATION PLZ

Kruka [31]

Answer:

1.25b

Step by step explanation:

6 0

3 years ago

A rectangular dartboard has an area of 648 square centimeters. The triangular part of the dartboard has an area of 162 square ce

never [62]

Answer:

The probability that the dart lands inside the triangle is 0.25

Step-by-step explanation:

* Lets explain how to find the probability of an event

- The probability of an Event = Number of favorable outcomes ÷ Total

number of possible outcomes

- P(A) = n(E) ÷ n(S) , where

# P(A) means finding the probability of an event A

# n(E) means the number of favorable outcomes of an event

# n(S) means set of all possible outcomes of an event

- P(A) < 1

* Lets solve the problem

- A rectangular dartboard has an area of 648 cm²

- The triangular part of the dartboard has an area of 162 cm²

- A dart is randomly thrown at the dartboard

- The dart lands in the rectangle

∴ The area of the rectangle is the set of all possible outcomes n(S)

- The probability P(A) that the dart lands inside the triangle

∴ The area of the triangle is set of favorable outcomes of an

event n(E)

∵ P(A) = n(E) ÷ n(S)

∴ P(T) = area of the triangle ÷ area of the rectangle

∵ Area of the rectangle is 648 cm²

∴ n(S) = 648

∵ The area of the triangle is 162 cm²

∴ n(E) = 162

∴ P(T) = 162 ÷ 648 = 1/4 = 0.25

* The probability that the dart lands inside the triangle is 0.25

5 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