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3 years ago

What is the value of “three less than the quotient of six and a number, increased by nine” when n = 3?

Mathematics

2 answers:

iVinArrow [24]3 years ago

7 0

Answer:

Step-by-step explanation:

-3 + 6/n + 9

for n=3 is

-3 + 6/3 + 9

= -3 +2 + 9

= -1 + 9

= 8

podryga [215]3 years ago

7 0

Answer:

Step-by-step explanation:

three less than the quotient of six and a number, increased by nine”

we write the given statement in expression form

Let the unknown number be 'n'

quotient of six and a number is 6/n (for quotient we divide)

three less than 6/n means we need to subtract 3 from 6/n

6/n - 3

Finally, it is increased by 9

So expression becomes $\frac{6}{n} -3 +9$

find the value of the expression when n=3

Plug in 3 for n

$\frac{6}{3} -3 +9$

2 -3 +9 = 8

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Suppose you carry out a significance test of h0: μ = 8 versus ha: μ > 8 based on sample size n = 25 and obtain t = 2. 15. Fin

OLEGan [10]

Using a t-distribution calculator and finding the p-value, the correct option regarding the conclusion is given by:

a) the p-value is 0.02. We reject h0 at the 5% significance level because the p-value 0.02 is less than 0.05.

<h3>What is the relation between the p-value and the conclusion?</h3>

It also involves the significance level, as follows.

If the p-value is less than the significance level, we reject the null hypothesis $h_0$ .

If it is more, we do not reject.

In this problem, a t-distribution calculator for a right-tailed with <em>t = 2.15 and 25 - 1 = 24 df</em> is used to find a p-value of 0.02.

It is less than 0.05, hence option a is correct.

More can be learned about p-values at brainly.com/question/26454209

5 0

2 years ago

Help pls and no links

Svetlanka [38]

We know a or b != 0

but a-b==0 so both are the same unknown number

b/a must = 1

a/b is equivalent to b/a as a/b=1 as well.

So the last answer choice.

7 0

3 years ago

Cqn someone answer this?

Norma-Jean [14]

Answer:

-3

Step-by-step explanation:

-11/2+27/4+(-17/4)
(-22+27-17)/4
(-12)/4
-3

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3 years ago

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2. A bag contains 3 red, 4 white, and 5 blue marbles. Jason begins removing marbles from the bag at

galben [10]

Answer:

the total number of marbles are 12 and there are 3 colours so you divide 12 by 3

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3 years ago

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Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

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4 years ago