All categories

3 years ago

I need answers plz help me

Mathematics

1 answer:

ddd [48]3 years ago

5 0

The answer is C.
I added the fractions of the leftovers together and I got 25/24.

You might be interested in

The equation for Y=x+5 and y=-2x-1

Olenka [21]

It’s y=3 i’m pretty sure

8 0

3 years ago

PLEASE HELP I WILL GIVE 50 POINTS!

Anastaziya [24]

Answer:

Domain: x is all real numbers

Range: y is less than or equal to 5

3 0

4 years ago

A) Use your calculator to work out 182 - 7.9

riadik2000 [5.3K]

Give me brainiest please

8 0

3 years ago

Need help on this question please

Natasha_Volkova [10]

Hello!

This question is an example of function composition.

Given, f(x) = 3x - 5, find f(x - 1). For this problem, we substitute x for (x - 1).

f(x - 1) = 3(x - 1) - 5

f(x - 1) = 3x - 3 - 5

f(x - 1) = 3x - 8

Therefore, the answer is choice D, f(x - 1) = 3x - 8.

5 0

3 years ago

A stack of cards consists of seven red and four blue cards. A second stack of cards consists of eleven red cards. A stack is sel

Ann [662]

Answer:

0.1733 = 17.33% probability the first stack was selected.

Step-by-step explanation:

To solve this question, it is needed to understand conditional probability, and the hypergeometric distribution.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Conditional probability:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Probability of all red cards for the first stack:

For this, we use the hypergeometric distribution, as the cards all chosen without replacement.

7 + 4 = 11 cards, which means that $N = 11$ .

7 red, which means that $k = 7$

3 are chosen, which means that $n = 3$

We want all red, so we find P(X = 3).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,11,3,7) = \frac{C_{7,3}*C_{4,0}}{C_{11,3}} = 0.2121$

Conditional probability:

Event A: All red

Event B: From the first stack.

Probability of all red cards:

0.2121 of 50%(first stack)

1 of 50%(second stack). So

$P(A) = 0.2121*0.5 + 1*0.5 = 0.60605$

Probability of all red cards and from the first stack:

0.21 of 0.5. So

$P(A \cap B) = 0.21*0.5 = 0.105$

What is the probability the first stack was selected?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.105}{0.60605} = 0.1733$

0.1733 = 17.33% probability the first stack was selected.

4 0

3 years ago