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4 years ago

Evaluate. -|-2 + 9| i need help

1 answer:

amm18124 years ago

8 0

Hey there!!

Okay so let's start simplifying this :D

-| -2 + 9| <---- remember that the "vertical bars" are absolute value signs

Absolute value signs will make any values within them into a positive value even if it was negative in the first place

-| 7 | <---- first simplify what is within the absolute value signs

<u>- 7</u><u /> <----- this is your final answer!!!

I hope this helps you! ^__^

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Think about a situation when you needed to analyze data. What types of trends did you find in the data? How did noticing the tre

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3 years ago

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

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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:

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

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3 years ago

What can be useful when finding the value of a variable

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Writing an equation would be most helpful, but depending on the situation drawing a diagram or reading a table could work better.

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3 years ago

Drag each set of column entries to the correct location in the matrix equation. Not all sets of entries will be used. A biker ne

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Answer:

The matrix equation is $\left[\begin{array}{ccc}1&1&1\\1&-1/2&0\\0&2&-1\end{array}\right]=\left[\begin{array}{c}120&35&20\end{array}\right]$

Step-by-step explanation:

* Lets change the story problem to equations

- The distance between the starting point and checkpoint 1 is x

- The distance between checkpoint 1 to checkpoint 2 is y

- The distance between checkpoint and the finish line is z

- The total distance for the race is 120 miles

∴ x + y + z = 120 ⇒ (1)

-The distance from the starting point to checkpoint 1 is 35 miles

more than half the distance from checkpoint 1 to checkpoint 2

∵ The distance from the starting point to checkpoint 1 is x

∵ The distance from checkpoint 1 to checkpoint 2 is y

- x is more than half y by 35

∴ x = 35 + (1/2) y ⇒ subtract (1/2) y from both sides

∴ x - (1/2) y = 35 ⇒ (2)

- The distance from checkpoint 2 to the finish line is 20 miles less

than twice the distance from checkpoint 1 to checkpoint 2

∵ The distance from checkpoint 2 to the finish line is z

∵ the distance from checkpoint 1 to checkpoint 2 is y

- z is less than twice y by 20

∴ z = 2y - 20 ⇒ add 20 to both sides

∴ z + 20 = 2y ⇒ subtract z from both sides

∴ 2y - z = 20 ⇒ (3)

* Now lets write the three equations

# x + y + z = 120 ⇒ (1)

# x - (1/2) y = 35 ⇒ (2)

# 2y - z = 20 ⇒ (3)

- Now lets write the matrix equation that models this situation

∴ $\left[\begin{array}{ccc}1&1&1\\1&-1/2&0\\0&2&-1\end{array}\right]=\left[\begin{array}{c}120&35&20\end{array}\right]$

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3 years ago

Read 2 more answers

Hana is playing a virtual reality game in which she must toss a disc to land on the largest triangular section of the board. If

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Answer:

Her probability of success is 1/2

Step-by-step explanation:

Firstly, we have to calculate the area of the entire board

That would be the rectangle of the measure 50 cm by 40 cm

We have the area as the product of this two

That would be 50 * 40 = 2,000 cm^2

Now, the value calculated is the area of the entire possible landing area

Secondly, we need to get the area that counts as a success

that would be the area of the triangle that has a base of 50 cm and a height of 40 cm

The area of this is the product of the two divided by 2

we have this as ;

50 * 40 * 1/2 = 1,000 cm^2

So, her probability of success is simply the area of the triangle divided by the area of the rectangle

= 1000/2000 = 1/2

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3 years ago