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4 years ago

15

Which statement describes a unit rate? A) Sara ate 1 cookie B) Sara is driving 16 miles C) Sara is driving 30 miles per 1 hour D

) Sara ate 3 crackers & 1 apple

2 answers:

lapo4ka [179]4 years ago

8 0

A unit rate is a ratio with a number in comparison to 1 (per 1). So, In this case...

C) Sara is driving 30 miles per 1 hour is an example of unit rate.

VikaD [51]4 years ago

4 0

C is correct because it's saying that Sara is driving 30 miles per hour. Unit rate is basically a way of writing or talking about a ratio compared to a value of one and in this case, the value is 30 per hour (1 hour).

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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

Can someone help me with this problem? thank you!

abruzzese [7]

Answer:

When 6x - 4 = 8, x = 2.

Step-by-step explanation:

6x - 4 = 8

Add 4 to both sides.

6x = 12

Divide both sides by 6.

x = 2

Proof:

6x - 4 = 8

Substitute variable.

6(2) - 4 = 8

Multiply 6 and 2.

12 - 4 = 8

Subtract 4 from 12.

8 = 8

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2 years ago

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4 years ago

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Hey can you please help me posted picture of question

juin [17]

For this case we can have the following function:
r = f (t)
Where,
Independent variable is t.
Dependent variable is r.
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FALSE
option B

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4 years ago

Pls help me I am confused

Maslowich

Answer:

$\huge\boxed{A.\ \dfrac{565}{182}}$

Step-by-step explanation:

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3 years ago

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