All categories

3 years ago

You know the distance between two cities is 140 miles. What is the distance in kilometers?

2 answers:

7 0

The unknown is the distance in kilometres
The given is the distance in miles=140 miles
1 mile=1.6093km
140 miles=?
140/1*1.6093km=225.302km
Therefore the 140 miles in km is 225.302km

Marysya12 [62]3 years ago

6 0

Answer:

225.302 km.

Step-by-step explanation:

Given :The distance between two cities is 140 miles.

To Find: What is the distance in kilometers?

Solution:

The distance between two cities = 140 miles

1 mile = 1.6093 km

140 miles = $140 \times 1.6093$

= $225.302$

The distance between two cities in km = 225.302 km

Hence the distance in kilometers is 225.302 km.

So, Option C is true .

You might be interested in

Graph the inequality and check the solution.

denis-greek [22]

Answer:

B,C,E

Step-by-step explanation:

did the assignment on Edge 2020

B. draw an open circle at 100

C. shade all numbers to the right of 100

E. 150 is shaded, so substitute 150 in for the variable to check the graph of the solution

5 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

Somebody please help asapppppp

anzhelika [568]

The function was moved two units to the left (a)

8 0

3 years ago

2/3 * 10/24*-3 1/8* -4/5

nadya68 [22]

Answer: 25/36

Step-by-step explanation:

Multiply:

2/3 * 10/24 = 20/72 = 5/18

-3 1/8 * -4/5 = 5/2

5/18 * 5/2 = 25/36

7 0

2 years ago

For every 15 Cheetos I ate 300 calories what's the unit rate

gregori [183]

Answer:

20 calories per Cheetos

Step-by-step explanation:

hope it helped

5 0

3 years ago