If f(x) = 4 – x2 and g(x) = 6x, which expression is equivalent to (g

A solid machine part is to be manufactured as shown in the figure. The part is made by cutting a small cone off the top of a lar

Alexxx [7]

Answer:

The answer is 390 × pi.

Step-by-step explanation:

Big cone:

V = 1/3 x pi x r^2 x h

= 1/3 x 3.14 x 9^2 x 15

= 1272.3 cubic inches

Small cone:

V = 1/3 x pi x r^2 x h

= 1/3 x 3.14 x 3^2 x 5

= 47.1 cubic inches

1272.3 - 47.1 = 1225.2 cubic inches

1272/3.14 ≈ 390

The answer is 390 × pi.

4 0

1 year ago

You and your friend each collect rocks and fossils. Your friend collects three times as many rocks and half as many fossils as y

pochemuha

Answer:

I collect 4 rocks and 16 fossils. While my friend collects 12 rocks and 8 fossils.

Step-by-step explanation:

The number of total objects collected by me = 20

The number of objects collected by my friend = 20

Let us assume:

The number of rocks collected by me = m

The number of fossils collected by me = 2 k

⇒ m + 2 k = 20 .... (1)

So, according to the question:

The number of rocks collected by my friend = 3 x (rocks collected by me)

= 3 x (m) = 3 m

The number of fossils collected by my friend = 1/2 x (fossils collected by me) = 1/2 x (2k ) = k

⇒ 3 m + k = 20 .... (2)

Solving (1)and (2) for the value of m and k, we get:

m + 2 k = 20 ⇒ m = 20 - 2 k ... Put this in (2)

3 m + k = 20 ⇒ 3 (20 - 2 k) + k = 20

⇒ 60 - 6 k + k = 20

or, - 5k = -40

or, k = 40/5 = 8

⇒ k= 8

m + 2(8) = 20

or, m = 4

Hence, the number of rocks collected by me = m = 4

The number of fossils collected by me = 2 k = 2 (4) = 16

Hence, the number of rocks collected by my friend = 3 m = 3 x 4 = 12

The number of fossils collected by my friend = k = 8

Hence, I collect 4 rocks and 16 fossils. While my friend collects 12 rocks and 8 fossils.

4 0

1 year ago

Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

 $P(D)= \frac{1}{3}$ 

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

3 0

2 years ago

TJ is thinking of a number

Likurg_2 [28]

Answer:

$\frac{1}{3}n - 5$

Step-by-step explanation:

If n is the number that TJ is thinking, then says he found 1 third of the number, you would multiply 1/3 by n. Then he says to subtract 5, which would be the -5.

8 0

1 year ago

Which algebraic expressions are polynomials? Check all that apply. x2y 2 – 4x3 + 12y – x2 – 16 3.9x3 – 4.1x2 + 7.3

Softa [21]

Polynomials are the expressions that contain two or more algebra terms - constants, variables and their powers of a particular variable. The simple example of a polynomial is $2x^{2} + 3x^{3} + 5x + 2$

In the given question,
9x3 – 4.1x2 + 7.3 is the polynomial because it has different powers of same variable x and does not contain any other variable.

4 0

2 years ago

Read 2 more answers

If f(x) = 4 – x2 and g(x) = 6x, which expression is equivalent to (g – f)(3)?