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3 years ago

What number are 1.3 units from 0.4 on a number line

Mathematics

2 answers:

spayn [35]3 years ago

3 0

0.5 ,0.6 ,0.7 ,0.8 ,0.9 ,1,1 .1 ,1.2 ,1.3

m_a_m_a [10]3 years ago

3 0

Move 1.3 units to the right from 0.4

1.3 + 0.4 = 1.7

The number is 1.7

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Trail mix is sold in 1-pound bags. Mary will buy some trail mix And re-package it si that each f the 15 members of her hiking cl

mixer [17]

<span>Okay. so we need 15 2/5 pound bags. We'll calculate how many pounds that actually comes to. Our problem is 15x2/5. Using cross simplification, our problem becomes 3x2. 3x2=6. We need 6 pounds. That means that we must have 6 1-pound bags to fit with no leftovers. </span>

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3 years ago

Read 2 more answers

A certain lot consisting of ten items has three defective items and seven nondefective items. How many possible subsets of 2 ite

Charra [1.4K]

Answer:

Step-by-step explanation:

The order in which the items are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

How many possible subsets of 2 items can be chosen from this lot?

Combinations of 2 from a set of 10. So

$C_{10,2} = \frac{10!}{2!(10-2)!} = 45$

4 0

3 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago

If a number a is chosen at random from the set

Ivanshal [37]

Answer:

D. 3/19

Step-by-step explanation:

12/1 = 12

12/2 = 6

12/3 = 4

These are the only ones that make the equation true.

There are also only 19 numbers because they need to be less than 20 like the question indicated. Therefore, the answer is D.

5 0

3 years ago

What is the quadratic equation of the curve of best fit shown below

Kisachek [45]

First off, we know that the y-interecept is 2, so there is going to be a plus two in the final equation. Now with the two answers input the number 8 to see which will output 6.

6 = 1/12(8)^(2) + 2
6 = 1/12(64) + 2
6 = 64/12 + 2
6 = 16/3 + 6/3
6 does not equal 22/3

6 = 1/16(8)^(2) + 2
6 = 1/12(64) + 2
6 = 64/16 + 2
6 = 4 + 2
6 = 6
This means that y = 1/16 x^2 + 2 is the correct answer. I hope that helps!

4 0

3 years ago