All categories

3 years ago

1-4) What is the point of intersection?

Mathematics

1 answer:

Irina18 [472]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Find the volume of a cylinder with a diameter of 8 inches and a height that is three times the radius.

Eddi Din [679]

Step-by-step explanation:

volume = pi x radius² x height

okay so the cylinders diameter is 8 inches

diameter divided by 2 is radius so 8/2 = 4

so 3 times the radius will be 3x4 = 12

use the formula above and then you will get your answer

6 0

3 years ago

What linear equation is represented by the table? 1

rewona [7]

Answer:

Y=4x-2

Step-by-step explanation:

You have to try it (x,y) (2,6)

y=4x-2

6=4(2)-2

6=8-2

8 0

3 years ago

An artist cuts 4 squares with side length x ft from the corners of a 12 ft-by-18 ft rectangular piece of sheet metal. She bends

Kruka [31]

Answer:

$V=4x^3-60x^2+216x$

Step-by-step explanation:

<u>Volume And Function s</u>

Geometry can usually be joined with algebra to express volumes as a function of some variable. The volume of a parallelepiped of dimensions a,b,c is

$V=abc$

Our problem consists in computing the volume of a box made with some sheet of metal 12 ft by 18 ft. The four corners are cut by a square distance x as shown in the image below .

If the four corners are to be lifted and a box formed, the base of the box will have dimensions (12-2x)(18-2x) and the height will be x. The volume of the box is

$V=x(12-2x)(18-2x)$

Operating and simplifying

$V=4x^3-60x^2+216x$

8 0

3 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago

I need an answer for the empty spaces

abruzzese [7]

Answer:

4, +, 6

Step-by-step explanation:

3 0

3 years ago