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3 years ago

Judith used 11/14 of her tank of gas to drive 12 miles yesterday. How much gas did she use to drive one mile?

Mathematics

1 answer:

Digiron [165]3 years ago

3 0

Answer: she used 11/168 tank of gas per mile

Step-by-step explanation:

12 miles___11/14 tank of gas

1 mile____x= 11/168 tank of gas

(1 mile * 11/14 tank of gas)/12 Miles=

11/168 tank of gas

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In a school 40% are boys and 900 are girls. find the total number of students of the school and the number of boys

My name is Ann [436]

Answer:

1500 students

600 boys

Step-by-step explanation:

40% are boys, so 60% are girls.

Writing a proportion:

900 / 60% = x / 100%

x = 1500

There are 1500 students in the school, which means there are 600 boys in the school.

6 0

3 years ago

A student pulls a box of mass 5 kg with a force of

klemol [59]

Answer:

50j

Step-by-step explanation:

workdone = force x distance

workdone = 20x2.5

50j

7 0

3 years ago

Given: AB = BC Prove: B is the midpoint of AC. A line contains point A, B, C. A flow chart has 3 boxes that go from top to botto

aliina [53]

Answer:

flowchart proof

Step-by-step explanation:

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3 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

Multiply: 0.27055 · 1,000

photoshop1234 [79]

0.27055*1,000=

270.55

-Hunter

4 0

3 years ago

Read 2 more answers