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3 years ago

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A teacher and a doctor each have 45 books. If 4/5 f the teacher's books and 2/3 of the doctor's books are novels, how many more

novels does the teacher have than the doctor? Explain the process that was used to find your answer.

1 answer:

Anarel [89]3 years ago

8 0

Answer:

3/5

Step-by-step explanation:

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[ Let f ( x ) ] = [ 3 x 2 + 2 x ] , [ g ( x ) ] = [ 3 x + 9 ] , [ h ( x ) ] = [ 12 + x ] , [ What is f ( g ( 3 ) ) ? ] [ Let f (

Pachacha [2.7K]

Answer:

The value of f(g(3) = 1,008

Step-by-step explanation:

<u>Step:-1</u>

Given f(x) = $3 x ^ 2 + 2 x$

and g(x) = 3x+9

h(x) = 12+x

Given g(x) = 3x+9

put x =3

g(3) = 3(3)+9

<u> g(3) = 18</u>

<u>Step :-(2)</u>

f(g(3)) = f(18)

f(g(3) = 3(18)^2 +2(18) (since f(x) = $3 x ^ 2 + 2 x$ )

= 1,008

<u>Final answer</u>:-

The value of f(g(3) = 1,008

8 0

3 years ago

How do I solve and find x for this question

Fed [463]

Answer: -9.1

Explanation
-3x +12.4=39.7
First minus 12.4 from both sides

-3x +12.4=39.7
-12.4 -12.4
-3x=27.3
Now devid both sides by -3
X= -9.1

7 0

2 years ago

A shipment of 11 printers contains 2 that are defective. Find the probability that a sample of size 2, drawn from the 11, will

svet-max [94.6K]

The required probability is $\frac{36}{55}$

<u>Solution:</u>

Given, a shipment of 11 printers contains 2 that are defective.

We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.

Now, we know that, $\text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}$

Probability for first draw to be non-defective $=\frac{11-2}{11}=\frac{9}{11}$

(total printers = 11; total defective printers = 2)

Probability for second draw to be non defective $=\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}$

(printers after first slot = 10; total defective printers = 2)

Then, total probability $=\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}$

7 0

3 years ago

The equations of four lines are given. Identify which lines are perpendicular.

erma4kov [3.2K]

Answer:

Line 2 and Line 4 are perpendicular

Step-by-step explanation:

Line 2: y=1/5x−3

Line 4: y+1=−5(x+2)

3 0

3 years ago

51% of the tickets sold at a school carnival were early-admission tickets. If the school sold 100 tickets in all, how many early

saul85 [17]

Answer:

51 were the total number of the tickets sold at a school carnival were early-admission tickets.

Step-by-step explanation:

Total number of tickets sold by = 100

Let $x$ be the early-admission tickets sold by the school.

As 51% of the tickets sold at a school carnival were early-admission tickets.

so

$x=\frac{51}{100}\times 100$

$= 51$

Therefore, 51 were the total number of the tickets sold at a school carnival were early-admission tickets.

5 0

3 years ago