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3 years ago

What's the equation for q and r

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

8 0

5 + 5 = 10
10 + 5 = 15
20 + 5 = 25
40 + 5 = 45

q + 5 = r

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Part of a regression output is provided below. Some of the information has been omitted.

quester [9]

Answer:

301.189

Step-by-step explanation:

Given the table :

Source of Variation - - SS - - - df - - - MS - - - - F

Regression - - - - - -3177.17 - - -2 - - 1588.6

Residual - - - - - - - - ______ --17 - - -17.717

Total - - - - - - - - - - 3478.36 - - 19

Calculate the SSR, Sum of Square residual

The Sum of Square RESIDUAL (SSR), Mean Square Residual (MSR) and Degree of Freedom RESIDUAL (DFR) are related by the formular :

MSR = SSR / DFR

Hence,

SSR = MSR × DFR

Fr the table ;

MSR = 17.717 ; DFR = 17

SSR = (17.717 × 17)

SSR = 301.189

5 0

3 years ago

In one company, 20% of the workers in the technical department applied for vacation at the same time. If 12 workers applied for

FinnZ [79.3K]

Answer:

There are 60 tech workers altogether

Step-by-step explanation:

Let X be the total number of tech workers in the company

Out of X tech employees, 20 % applied for leaves

In totality, 12 tech employees are on leave

Thus,

$\frac{20}{100} * X = 12$

On solving the above mathematical equation, we get -

$X = \frac{12*100}{20} \\X = 60$

Hence, there are 60 tech workers altogether

6 0

2 years ago

A 6 sided number cube is tossed. what is the probability of rolling a 6 or an odd number?

Veronika [31]

For any probability the formula is $P_{desired}=\frac{desired}{total}$ . In this case we have 4 desired outcomes (the odds, 1, 3 and 5, and also 6) out of 6 total sides, so our probability $P_{oddor6} =\frac{odd or 6}{total sides}=\frac{4}{6} = \frac{2}{3}$ .

Mutually exclusive means the events cant occur at the same time. In our case it is not possible to roll both a 6 and an odd number at the same time since 6 is even. Therefore the events ARE mutually exclusive.

Combine the two parts and C is the correct answer. <span />

3 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Solve the inequality n + 20 < -42, and write the solution in interval notation.

Firdavs [7]

Answer:

Step-by-step explanation:

4 0

3 years ago