How many different simple random samples of size 5 can be obtained from a population whose size is 51​? The number of simple random samples which can be obtained is 281887200281887200. ​(Type a whole​ number.)

Answers

Answer 1
Answer:

Answer:

2349060

Step-by-step explanation:

Given that :

Population size (n)= 51

Number of simple random samples (r) = 5

Using the combination formula :

nCr = n! ÷ (n-r)! r!

51C5 = 51! ÷ (51 - 5)!5!

51C5 = 51! ÷ 46!5!

51C5 = (51*50*49*48*47)*46! / 46! * (5*4*3*2*1)

51C5 = (51*50*49*48*47) / (5*4*3*2*1)

51C5 = 281887200 / 120

= 2349060


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The distance d of a particle moving in a straight line is given by d(t) = 2t3 + 5t – 2, where t is given in seconds and d is measured in meters. Find an expression for the instantaneous velocity v(t) of the particle at any given point in time.

Answers

Answer:

6t^2 + 5

Step-by-step explanation:

i took the test and somehow got it correct:)

Answer:

Step-by-step explanation:

hello,

v(t)=(d(t))/(t)=2t^2+5-(2)/(t)

hope this helps

How does the quotient compare to the dividend when the divisor is less than 1

Answers

Final answer:

When a dividend is divided by a divisor that is less than 1, the resulting quotient is greater than the original dividend. This is equivalent to multiplying the dividend by the reciprocal of the divisor.

Explanation:

In mathematics, when we divide any number (the dividend) by a number that is less than 1 (the divisor), the quotient will be greater than the dividend. This is because dividing by a number less than 1 is equivalent to multiplying by its reciprocate which is more than 1. For example, let's consider 10 divided by 0.5 (which is less than 1); the quotient is 20, which is greater than the dividend (10). Therefore, in relation to your question, the quotient is larger than the dividend when the divisor is less than 1.

Learn more about Division here:

brainly.com/question/33969335

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Answer:

Greater provided the divisor is positive.

It would be more accurate and clearer to say that when dividing a number which is greater than zero by a number between 0 and 1, then the quotient is greater than the dividend.

If the divisor is negative and the dividend is positive, then the quotient is negative (and so less than the dividend).

Hector waters his lawn if it does not rain at least 1 1/2 inches each week. It has already rained 3/8 inch this week. Which inequality represents the number of inches of rain, r, still needed so Hector does not have to water his lawn this week?A. r ≥ 1 7/8


B. r > 1 7/8


C. r ≥ 1 1/8


D. r > 1 1/8

(The fractions are mixed numbers!!!!)

Answers

Answer:

the answer is C

Step-by-step explanation:

this is because he needs at least 1 1/2, it's already rained 3/8 inches, so to simplify the equation change 1 1/2 to 1 4/8 which is still equivalent to 1 1/2, then subtract 3/8 from 1 4/8 which will give you 1 1/8. now here is where it gets kind of tricky, both C and D seem to have 1 1/2, but the difference between them is the symbols > and ≥ are CRUCIAL!!!! (">" means that the number at the left is greater than the number at the right while "≥" means that the number at the left is greater than OR EQUAL TO the number at the right) so as you can see you also need to pay attention to what the text says, because it is mention quote on quote "Hector waters his lawn if it does not rain at least 1 1/2 inches each week," this means that if it's not at least (in other words the same as or more) 1 1/2 inches of rain a week, he will water his plants

Can you help me to do this please?Determine the inverse of the h(x)

Answers

Answer:

h^(-1)(x)=x^3+6x^2+12x+7

Explanation:

Given the below function;

h(x)=\sqrt[3]{x+1}-2

We'll follow the below steps to determine the inverse of the above function;

Step 1: Replace h(x) with y;

y=\sqrt[3]{x+1}-2

Step 2: Switch x and y;

x=\sqrt[3]{y+1}-2

Step 3: Solve for y by first adding 2 to both sides;

\begin{gathered} x+2=\sqrt[3]{y+1}-2+2 \n x+2=\sqrt[3]{y+1} \end{gathered}

Step 4: Take the cube of both sides;

\begin{gathered} (x+2)^3=(\sqrt[3]{y+1})^3 \n (x+2)^3=y+1 \end{gathered}

Step 5: Expand the cube power;

Recall;

(a+b)^3=a^3+3a^2b+3ab^2+b^3

Applying the above, we'll have;

\begin{gathered} (x+2)^3=y+1 \n x^3+3x^2\cdot2+3x\cdot2^2+2^3=y+1 \n x^3+6x^2+12x+8=y+1 \end{gathered}

Step 6: Subtract 1 from both sides of the equation;

\begin{gathered} x^3+6x^2+12x+8-1=y+1-1 \n x^3+6x^2+12x+7=y \n \therefore y=x^3+6x^2+12x+7 \end{gathered}

Step 7: Replace y with h^-1(x);

h^(-1)(x)=x^3+6x^2+12x+7

Prove that if a,b and c are odd integers such that a + b +c = 0, then abc < 0. (you are permitted to use well-known properties of integers here.) be sure to show all work.

Answers

The sum of three odd integers will never be an even integer (0).

This proof is impossible.

16x - 10y = 10
-8x - 6y = 6

Answers

Answer:

x=0 y=-1

Step-by-step explanation:

16x-10y=10

-8x-6y=6 *2  -16x-12y=12

16x-10y=10

-16x-12y=12

-22y=22 Add it together

y=-1

-8x-(6*(-1))=6

-8x+6=6

x=0