Please help!! Someone said it was d.... and I think it’s either b or c..... now I am very confused :( giving 12-15 points!!

What is the median of the following set of numbers? 33, 30, 29, 28, 27, 34, 31

IrinaVladis [17]

Median is a middle value
So it is 28

4 0

3 years ago

Caroline wants to buy a pair of shoes that has an original price of $40.00. She has a coupon for a 30% discount off of the origi

fredd [130]

Answer:
$26.32

step-by-step explanation:
it originally was $40.00
she has a coupon for 30%
30% of 40.00 is 12
40 • 0.30 = 12
40 - 12
$28
it is now $28 after the coupon
her state charges an 6% sales tax on the discounted price
6% of 28 is 1.68
28 • 0.06 = 1.68
28 - 1.68 = 26.32

6 0

2 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

American cars maker produce 5650000 cars each year in Europe that Americans make 6 million 550 cars the mistake did Ben make how

Vlad [161]

Answer:

Ben has mixed up the digit for millions and hundred thousands: he wrote 5 in the place for millions and 6 in the place for hundred thousands, but it should be the other way. I think that he should exchange those two digits in the faulty report.

3 0

2 years ago

A ball is thrown vertically upwards with an initial velocity of 22 m/s. When will its speed be 10

gogolik [260]

Answer:

<em>The ball's speed will be 10 m/s at t=1.22 seconds</em>

Step-by-step explanation:

The vertical motion of an object is controlled by the force of gravity. This means that there is a non-zero net force acting on the object that makes it accelerate downwards.

If the object is thrown upwards at speed vo, its speed at time t is:

$v_f=v_o-g.t$