I need help with 3 B please!

A variable needs to be eliminated to solve the system of equations below. Choose the

AURORKA [14]

Two different things u can do
Substitution:
X-9y=12
+9y +9y
X=9y+12

-3(9y+12)+9y=36
-27y-36+9y=36
+36 +36
-18y = 72
/-18 /-18
Y = -4

Or whatever this other I’ve is called

-3x-9y=36
-1(X-9y)=(12)-1

-3x-9y=36
-x+9y=-12

-4x=24
X= -6

8 0

3 years ago

Mr Gardner is making 6 treat bags. He has 185 chocolate-covered raisins to share evenly among the treat bags.

Amiraneli [1.4K]

Answer:

✎There will be 30 Chocolate-Covered raisins in each bag.

✎ And 5 Remaining.

Step-by-step explanation:

Take 185 and divide it by 6 and you should get 30 per bag with a remainder of 5 :)

6 0

3 years ago

Read 2 more answers

10% of what is 80,000?

Nataly [62]

Answer:

let the no be x ..then,

10% of x =80,000

10x=80,000*100

x=80,00,000/10

x=8,00,000

hence 10%of 8,00,000 is 80,000

6 0

3 years ago

Read 2 more answers

Find 601st term of the arithmetic sequence whose first term is -85 and the common difference is 1/6.

irakobra [83]

Answer:

Step-by-step explanation:

Arithmetic sequence

Tn = a+(n-1)d

= -85 +(601-1) 1/6

= -85 + 600/6

= -85 + 100

= 15

601th term is 15

8 0

3 years ago

A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra

VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

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.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that $N = 6$

4 blue, which means that $k = 4$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

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

$P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4$

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that $N = 6$

2 yellow, which means that $k = 2$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

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

$P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667$

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that $P(A) = 0.4 + 0.0667 = 0.4667$

Probability of both being the same color and blue:

40% probability that both are blue, which means that $P(A \cap B) = 0.4$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571$

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

8 0

3 years ago