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2 years ago

10 points Help me with all I mark you as brainlist please

Mathematics

1 answer:

san4es73 [151]2 years ago

5 0

Answer: Question 1

a)5.47

b)4.53

Step-by-step explanation:

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Divide mentally. 6.2 ÷ 2,000 = ? A. 0.31 B. 0.031 C. 0.0031 D. 0.00031

soldi70 [24.7K]

C. 0.0031

6.2 divided by 2,000 = 0.0031

5 0

2 years ago

Randy has 128 ounces of dog food.he feeds his dog 8ounces of food each day.how many days will the dog food last?

pentagon [3]

The dog food food will last for 16 days
16x8=128

6 0

2 years ago

Read 2 more answers

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

PLEASE HELP PLEASEEEEE

antiseptic1488 [7]

Answer: 7.48

Step-by-step explanation:

Linear graph. take the point at 3 and multiply it by 2

3.74 x 2 = 7.48

5 0

2 years ago

Finding measurements of angles

Rom4ik [11]

We will be using the Angle Addition Postulate in this problem.

This states that if <em>c</em> is on the interior

of <ABD, then m<ABC + m<CBD = m<ABD.

Since m<ABC is 4x + 2, m<CBD is 3x - 7, and m<AVD is 100,

our equation will be 4x + 2 + 3x - 7 = 100.

Solving from here, we first simplify the left side to get 7x - 5 = 100.

Now add 5 to both sides to get 7x = 105.

Dividing both sides by 7, we find that <em>x = 15</em>.

Now we can use the value of x to help us find m<ABC.

Since the m<ABC is 4x + 2, we can substitute a 5 in for x.

This gives us 4(15) + 2 or 60 + 2 which is 62.

So m<ABC is 62°

8 0

2 years ago