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4 years ago

5

A person eating at a cafeteria must choose 4 of the 11 vegetables on offer. Calculate the number of elements in the sample space

for this experiment.

1 answer:

Charra [1.4K]4 years ago

4 0

Answer:

330

Step-by-step explanation:

11C4 = 330

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The supermarket has pie filling

kondor19780726 [428]

Answer:

$5.24

Step-by-step explanation:

If Mom bought 4, she bought $$1.19*4=$4.76$ worth of pie filling. Since she paid $10, she needs $$10-$4.76=$5.24$ change

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3 years ago

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The equation for Constant of Proportionality is y = kx? If you solved for the letter k instead of y, would the correct equation.

telo118 [61]

I don’t think it would work it you solved for k instead of y. Because you’re solving for y when you do constant of proportionality. i’m not sure if i’m right or not but that’s what i think.

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3 years ago

There are 135 fifth-grade students in a certain school. Each table on the lunchroom seats six students. How many tables are need

choli [55]

Answer:you just need to divided 135÷ 6

Step-by-step explanation:

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4 years ago

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For her birthday, Monica can invite 6 of her 15 student friends to join her at a theme park. If she chooses to invite friends at

svlad2 [7]

Answer:

B. 1 out of 5,005

Step-by-step explanation:

Given

Number of Friends = 15

Required

Probability of selecting 6 friends

The first step is to calculate the number of ways 6 friends can be selected

The keyword in the above statement is selection;

This implies combination;

The number of ways is calculated as follows;

$\left[\begin{array}{c}n&r&\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 15 and r = 6

$\left[\begin{array}{c}n&r&\end{array}\right] = \frac{n!}{(n-r)!r!}$

becomes

$\left[\begin{array}{c}15&6&\end{array}\right] = \frac{15!}{(15-6)!6!}$

$\left[\begin{array}{c}15&6&\end{array}\right] = \frac{15!}{9!6!}$

$\left[\begin{array}{c}15&6&\end{array}\right] = \frac{15 * 14 * 13 * 12 * 11 *10 * 9!}{9! *6 * 5 * 4 * 3 * 2 * 1}$

$\left[\begin{array}{c}15&6&\end{array}\right] = \frac{15 * 14 * 13 * 12 * 11 *10}{6 * 5 * 4 * 3 * 2 * 1}$

$\left[\begin{array}{c}15&6&\end{array}\right] = \frac{3603600}{720}$

$\left[\begin{array}{c}15&6&\end{array}\right] =5005$

Hence, there are 5005 ways of selecting 6 from 15 friends

Since, there's only one way of selecting the 6 named friends

Then, the probability is 1 out of 5,005

3 0

3 years ago

Using the definition of Covariance, Cov(X,Y) = E[(x – My)(Y – My)], prove the followings. a. Cov(X,Y) = E(XY) - Mx Hy b. Cov(X,Y

exis [7]

Answer:

Step-by-step explanation:

We have by definition

$Cov(X,Y) = E[(x – M_x)(Y – M_y)]$

where Mx and My are means of X and Y respectively

a) $E[(x – M_x)(Y – M_y)]\\= E(x,y) - E(x,My)-E(y,Mx)+M_x M_y\\=E(x,y) -E(x) M_y -E(y) M_x +M_x M_y\\=E(x,y)-M_x M_y-M_x M_y+M_x M_y\\=E(x,y)-M_x M_y$

b) Since right side inside expectation is commutative, we get cov(x,y) = cov (y,x)

c) $cov (x,x) = E(x-M_x)(x-M_x) = E(x-M_x)^2\\= Var(X)$

d) $Cov(X + Z,Y)=E(x+z-M_{x+z} )(Z-M_z)\\=E{(x-M_x)+(Y-M_y)}(Z-M_z)\\\\=E{(x-M_x)}(Z-M_z)+E{+(Y-M_y)}(Z-M_z)= cov (x,y)+cov (z,y)$

f) Var(x+y) = $E{(x-M_x)+(Y-M_y)}^2$

Expanding inside we get

= $Var(x)+var(y)+2 cov (x,y)$

8 0

3 years ago