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3 years ago

Find the greatest common factor of the following monomial 6bc and 9b^2c

Mathematics

2 answers:

galina1969 [7]3 years ago

6 0

Both 6bc and 9b²c are multiples of the number 3bc - we can rewrite them as

3bc · 2 and 3bc · 3b

Since the two monomials share the factors 3, b and c, their greatest common factor is 3bc.

lions [1.4K]3 years ago

4 0

The greatest common factors would be the biggest number that goes into both of these monomial.

As we can see, they both have one b and one c. This is the largest number of b’s and c’s it can have, as it can’t be any bigger.

Also, the greatest common factor of 6 and 9 is 3.

So, the answer is 3bc.

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In 2002, the Centers for Disease Control and Prevention (CDC) reported that 8% of women married for the first time by their 18th

ehidna [41]

Question:

In 2002, the Centers for Disease Control and Prevention (CDC) reported that 8% of women married for the first time by their 18th birthday, 25% married by their 20th birthday, and 76% married by their 30th birthday. Based on these data, what is the probability that in a family with two daughters, the first and second daughter will be married by the following ages? (Enter your answers to four decimal places.) (a) 18 years of age (b) 20 years of age (c) 30 years of age

Answer:

A.) 0.0064

B.) 0.0625

C.) 0.5776

Step-by-step explanation:

Given the following :

Married by 18th birthday = 8% = 0.08

Married by 20th birthday = 25% = 0.25

Married by 30th birthday = 76% = 0.76

In a family with two(2) daughters :

P(First daughter will be married by 18) = 0.08

P(second daughter will be married by 18) = 0.08

P(1st and 2nd married by 18) = (0.08×0.08) = 0.0064

B.)

P(First daughter will be married by 20) = 0.25

P(second daughter will be married by 20) = 0.25

P(1st and 2nd married by 20) = (0.25×0.25) = 0.0625

C.)

P(First daughter will be married by 30) = 0.76

P(second daughter will be married by 30) = 0.76

P(1st and 2nd married by 30) = (0.76×0.76) = 0.5776

6 0

2 years ago

Please help me with this !!! Thank you

puteri [66]

Answer:

messages me i will explain how to solve if i can

8 0

3 years ago

Solve the following inequality.<br><br> 9x−2>43

kirill [66]

Answer:

x > 5

Step-by-step explanation:

Given

9x - 2 > 43 ( add 2 to both sides )

9x > 45 ( divide both sides by 9 )

x > 5

8 0

3 years ago

PLEASE HELP THANKS YOU

kondaur [170]

U must find the median, the range, and the mode. And that is your answer

8 0

2 years ago

Read 2 more answers

A clock was reading the time accurately on Friday at noon. On Monday at 6pm the clock was running late by 468 seconds. On averag

Setler [38]

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × $\frac{1}{2}$ = 24

Saturday = 48

Sunday = 48

Monday = 48 × $\frac{3}{4}$ = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

To learn more about clock problems visit:

brainly.com/question/27122093.

#SPJ1

3 0

2 years ago