Construct the graph of the direct proportion y=kx for

A sphere with a radius of 4.8 centimeters is carved out of a right cone with a base radius of 8 centimeters and a height of 15 c

emmainna [20.7K]

Volume of the cone = 1/3 pi r^2 h = 1005.31 cm^3

Volume of the sphere = 4/3 pi r^3 = 463.25 cm^3

Volume of remaining portion of the cone = 1005.31 - 463.25

= 542.1 cm^3 to nearest tenth

4 0

3 years ago

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How? Just how I'm horrible at math

Oksanka [162]

Answer:

3 0

3 years ago

Please help!! I need the answer to the question in the picture!!!!

zloy xaker [14]

Answer:

$\large\boxed{D.\ x=-2(y+3)^2-1}$

Step-by-step explanation:

The equation of a horizontal parabola in the standard form:

$x=a(y-k)^2+h$

(h, k) - vertex

if a > 0, then open right

if a < 0, then open left

We have the vertrex (-1, -3) and the parabola is open left (a < 0).

Therefore the equation of a given parabola could be:

$x=-2(y-(-3))^2+(-1)=-2(y+3)^2-1$

8 0

3 years ago

In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 g

KIM [24]

Answer:

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

All girls from the first group:

20 students, so $N = 20$

10 girls, so $k = 10$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163$

All girls from the second group:

20 students, so $N = 20$

5 girls, so $k = 5$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006$

All girls from the third group:

20 students, so $N = 20$

8 girls, so $k = 8$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036$

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

$P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}$

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

7 0

3 years ago

ALGEBRA 1:

inna [77]

Distribution method
(2x+6)(2x-6)
=4x^2-12x+12x-36
=4x^2-36

7 0

3 years ago

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