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3 years ago

Which quantity could go in the blank to make the equation below true r+3r+_=6r

Mathematics

2 answers:

AveGali [126]3 years ago

5 0

Answer:

Step-by-step explanation:

r+3r_=6r

The left side would have to equal 6r

right now it is 4r

by adding 2r it would equal 6r making the equation true.

Hope this helps :)

creativ13 [48]3 years ago

4 0

Answer:

Step-by-step explanation:

r+3r+_=6r

Let's simplify this first. Subtract (r + 3r), or 4r, from both sides, obtaining

0 + _ = 2r

Then the quantity 2r, when placed in the blank, makes the equation true.

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g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa

adoni [48]

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is $\sigma = 0.0133$

Step-by-step explanation:

From the question we are told that

The upper specification is $USL = 1.68 \ mm$

The lower specification is $LSL = 1.52 \ mm$

The sample mean is $\mu = 1.6 \ mm$

The standard deviation is $\sigma = 0.03 \ mm$

Generally the capability index in mathematically represented as

$Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]$

Now what min means is that the value of CPk is the minimum between the value is the bracket

substituting value given in the question

$Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]$

=> $Cpk = min[ 0.88 , 0.88 ]$

$Cpk = 0.88$

Now from the question we are asked to evaluated the value of standard deviation that will produce a capability index of 2

Now let assuming that

$\frac{\mu - LSL }{ 3 * \sigma } = 2$

$\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2$

=> $0.08 = 6 \sigma$

=> $\sigma = 0.0133$

$\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }$

=> $2$

Hence

$Cpk = min[ 2, 2 ]$

$Cpk = 2$

So $\sigma = 0.0133$ is the value of standard deviation required

3 0

3 years ago

Use the variable d to represent the unknown quantity. the product of 4 and the depth of the pool

natka813 [3]

This can be represented by the expression 4d.

6 0

2 years ago

Anyone know how to solve this??

Tom [10]

$Given : \frac{(x^\frac{2}{5})^9\;.\;(x^\frac{-4}{15})}{x^\frac{1}{3}}$

$\implies {(x^\frac{18}{5})\;.\; (x^\frac{-4}{15})}\;.\;(x^\frac{-1}{3})$

$\implies x^(^\frac{18}{5} ^-^\frac{4}{15}^-^\frac{1}{3}^)$

$\implies x^(^(^\frac{54 - 4}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{50}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{10}{3}^)^-^\frac{1}{3}^) = x^(^\frac{9}{3}^) = x^3$

$\implies x^k = x^3$

$\implies k = 3$

7 0

3 years ago

Geometry, any idea?

Pepsi [2]

Answer:

Step-by-step explanation:

15x-31+9×+11+×=180

25×-20=180

25×=200

×=8

15(8)-31=89

9(8)+11=83

5 0

3 years ago

Please do this question

crimeas [40]

It would be 16 more 6th graders that are not in either band or orchestra.
15 6th Graders in band
9 6th Graders in both classes
+18 6th Graders in orchestra
-----
42 6th Graders in either class

100 6th Graders in total
-42 6th Graders in either class
-------
58 6th Graders in neither class
58-42= 16 more 6th Graders that are not in either class
You're welcome :)

7 0

3 years ago