Ask question

Ask question

All categories

3 years ago

13

I need help and it’s question 36 please answer

1 answer:

Eva8 [605]3 years ago

6 0

Answer:

try putting an imaginary 1 in front of your variable (c). so

1c + 4c = 15

combine like terms

(1c + 4c) = 5c

therefore we have

5c = 15

now, divide BOTH sides of the equation by 5

5c/5 = 15/5

5c/5 = c

15/5 = 3

c=3

You might be interested in

Jake wanted to measure the height of the Great Sphinx of Giza. He placed a mirror on the ground and then walked backwards until

irina1246 [14]

Jake wanted to measure the height of the Great Sphinx of Giza. He placed a mirror on the ground and then walked backwards until he was able to see the top <span>of the statue in the mirror. If his eyes were 5.5 feet above the ground, the statue is 87 ft. The answer is letter B.</span>

7 0

4 years ago

Write 37/100 as a decimal

kirza4 [7]

37 divided by 100 is .37 you move the . over 1 number for every 0

4 0

3 years ago

Read 2 more answers

Which equation is the correct translation of the following statement?

Verizon [17]

That is the last option

x - 5.75 = 10

7 0

4 years ago

If f(3)=2x^3 -2 , what is the value of f(2)

iris [78.8K]

I'm not sure about this but will give it a try:

Let f(n) = 2xⁿ - 2

Then f(3) = 2x³ - 2

So, f(2) = 2x² - 2

5 0

3 years ago

10 marbles what is the probability of drawing 2 blue marbles if the first marble is placed back in the bag before the second dra

stealth61 [152]

Complete Question:

There are 10 marbles. What is the probability of drawing 2 blue marbles if the first marble is NOT placed back in the bag before the second draw.

2 yellow, 3 pink, 5 blue

Answer:

$P(B_1\ and\ B_2) = \frac{2}{9}$

Step-by-step explanation:

Given

$Blue = 5$

$Pink = 3$

$Yellow = 2$

$Total = 10$

When the first blue marble is selected, the probability is:

$P(B_1) = \frac{n(Blue)}{Total}$

$P(B_1) = \frac{5}{10}$

Since the first marble is not returned, we have the following marbles left

$Blue = 4$

$Pink = 3$

$Yellow = 2$

$Total = 9$

The probability of picking a second blue marble is:

$P(B_2) = \frac{n(Blue)}{Total}$

$P(B_1) = \frac{4}{9}$

So, the required probability is:

$P(B_1\ and\ B_2) = P(B_1) * P(B_2)$

$P(B_1\ and\ B_2) = \frac{5}{10} * \frac{4}{9}$

$P(B_1\ and\ B_2) = \frac{5*4}{10*9}$

$P(B_1\ and\ B_2) = \frac{20}{90}$

$P(B_1\ and\ B_2) = \frac{2}{9}$

5 0

3 years ago