Ask question

Ask question

All categories

3 years ago

7

There are 12 cars contesting a race. The first three cars completing the race will be awarded prizes. In how many ways can the p

rizes be awarded?

1 answer:

Sergio039 [100]3 years ago

5 0

There would be 1,320 different ways, just multiply all the possible front runners 12 second runners 11 and third runners 10, 12x11x10=1,320

You might be interested in

a college freshman started the year with $4000 in spending money and after three months,she has $2800 left.assume that she will

gulaghasi [49]

Let's find out how much she spent every month.

4000 (starting money) - 2800 (remaining money) = 1200 spent over 3 months

1200/3 = 400 per month was spent

So if she continues to spend 400 a month?

How many months are left? 12 (months of the year) - 3 (months she already spent) = 9

So 9 (remaining months) * 400 (amt per month) = 3600 she'll spend at the going rate over 9 months.

But she only has 2800 left.

2800 (remaining) - 3600 (estimated total of spending) = -800

So she will be 800$ in debt at the end of the year at the current rate.

3 0

3 years ago

Ill mark brainlist plss help

Brilliant_brown [7]

It’s B! You multiply the height by width

6 0

3 years ago

Please help quickly

labwork [276]

I believe it is twenty eight. Ten is two thirds of fifteen. Twenty eight is two thirds of forty two. But I'm just assuming tbh

6 0

3 years ago

hey brainly users i would appreciate some help on making and solving a elimination problem, "There are 46 first graders on a fie

EleoNora [17]

Hopefully this is correct and what you wanted.

6 0

3 years ago

Verify that the following function is a probability mass function, and determine the requested probabilities. f left-parenthesis

Licemer1 [7]

Answer:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

Step-by-step explanation:

For this case we have the following density function:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

In order to satisfty that this function is a probability mass function we need to check two conditions:

1) $f(x_i) \geq 0, \forall x_i$

2) $sum_{i=1}^n P(X_i) =1$

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

And if we want to find the following probabilities:

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

7 0

3 years ago