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3 years ago

Which expression represents the number of possible combinations of n items taken rat a time

Mathematics

1 answer:

Likurg_2 [28]3 years ago

8 0

Answer:

nCr = n! / (r!)(n-r!).

Step-by-step explanation:

That would be nCr = n! / (r!)(n-r!).

A quick way of working these out is as follows:

Take 7C4:

this is 7*6*5*4/4*3*2*1 Cancelling out the 4 and 3*2:

= 7*5

= 35.

15C12 = 15C(15-12)

= 15C3

= 15*14*13 / 3*2*1

= 5*7*13

= 455.

You might be interested in

Can y’all help me on question 19

Zolol [24]

C is your answer. Because 8 times 1.75$ = $14.00 and 5 times 1.25= $6.25. So $14.00+ $6.25 = $20.25

6 0

3 years ago

A sample of 10 adult elephants had an average weight of 12,556 pounds. The standard deviation

kirill [66]

Answer:

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.25

The margin of error is:

M = T*s = 3.25*25 = 81

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12,556 - 81 = 12,475 pounds

The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.

The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.

6 0

3 years ago

An instructor wants to write a test with 25 questions where each question is worth 3, 4, or 5 points based on difficulty. He wan

irakobra [83]

ANSWER

Find out the how many 3, 4, and 5 point questions could there be.

To proof

Let us assume that the 3 points based question be = x

Let us assume that the 4 points based question be = y

Let us assume that the 5 points based question be = z

As given

An instructor wants to write a test with 25 questions

than the equation become in the form

x + y + z = 25

he quiz to be worth a total of 100 points.

than the equation is becomes

3x + 4y + 5z =100

As given

He wants the number of 3-point questions to be 4 less than the number of 4-point questions

x = y -4

Than the three equation are

x + y + z = 25 ,3x + 4y + 5z =100 and x = y -4

put x = y -4 in the x + y + z = 25 ,3x + 4y + 5z =100

than

2y + z = 29, 7y +5z= 112

multiply 2y + z = 29 by 5 and subtracted 7y +5z= 112

10 y -7y + 5y -5y = 145 -112

3y = 33

$y = \frac{33}{3}$

y =11

put in the x = y -4

x = 11-4

x= 7

put the value of x ,y in the x + y + z = 25

7 + 11 +z =25

18 +z = 25

z = 25 -18

z=7

therefore

numbers of the 3 point question be = 7

numbers of the 4 point question be = 11

numbers of the 5 point question be = 7

Hence proved

3 0

3 years ago

This is for geometry & I need help.

KiRa [710]

Answer:

1540 cubic millimeters

6 0

3 years ago

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

3 years ago