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2 years ago

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For a pyramid S.ABC of volume V. Let M be a point on side AB such that AM/AB = x. The plane (alpha) through M is parallel to two

straight lines SA, BC. the plane (alpha) divides the pyramid into 2 parts, in which the part containing the point B has a volume of V' with V' = 208/343V. The sum of the values of x satisfies the problem.
A. 3/2 B.3/7 C.135/686 D.0

2 answers:

Doss [256]2 years ago

4 0

Answer:

Can't understand your question

kondaur [170]2 years ago

4 0

Answer:

Hi...Have a wonderful day...

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Answer ASAP 15 points

malfutka [58]

A or 5/7, they’re both odd numbers so it cannot be simplified any further

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2 years ago

A cylindrical cardboard mailing tube has a diameter of 5 in. and a height of 30 in. A total of 30% of the tube’s capacity is fil

olga2289 [7]

First is to get the volume of the cylinder.
Volume = pi * r^2 * H
Volume = 3.1416 * (5/2)^2 * 30
Volume = 589.05 in^3

30% is already filled.
Filled Space = 589.05 * 0.30
Filled Space = 176.715 in^3
Empty Space = 412.335 in^3

Next, get the volume of the sphere.
V = (4/3)*pi*r^3
V = (4/3)*pi*(0.6/2)^3
V = 0.1130976 in^3

Number of foams = Filled Space / V
Number of foams = 176.715 / <span>0.1130976
Number of foams = 1562 Foams</span>

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3 years ago

Ms. Lanigan rolls a fair die 300

babymother [125]

Should be around 150 as 3/6 of the numbers on a die are even so thats 1/2 and 300x(1/5) or just 300x0.5 or just 300/2 equals to 150

5 0

2 years ago

Read 2 more answers

A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb

Maksim231197 [3]

Answer:

a) $P(X = 1) = 0.38742$

b) $P(X = 3) = 0.05740$

c) $P(X = 9) = 0.00000$

d) $P(X \geq 5) = 0.00163$

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

In this problem

There are 10 containers, so $n = 10$ .

A food-packaging apparatus underfills 10% of the containers, so $p = 0.1$ .

a) This is P(X = 1)

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

b) This is P(X = 3)

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

c) This is P(X = 9)

$P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000$

d) This is $P(X \geq 5)$ .

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

$P(X < 5) + P(X \geq 5) = 1$

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868$

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

$P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937$

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

$P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742$

So

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837$

Finally

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163$

3 0

2 years ago

Select the representation that does not change the location of the given point. (4, 110°)

miss Akunina [59]

Answer:

(4,470°)

Step-by-step explanation:

The representation that does not change the location of (4, 110°) in polar coordinates are all points that are coterminal with the given point.

The only point among the given options that is coterminal with (4, 110°) is (4, 470)°

The two points have the same magnitude and 470°-360°=110°.

Since 110° is coterminal with 470° and the two points have the same magnitude with the same sign, the two points represent the same location in polar coordinates.

The correct choice is (4,470°)

3 0

3 years ago