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shtirl [24]
3 years ago
12

Based on your results in Question 1, how is the volume of the cone related to the volume of the cylinder, given that their bases

and heights are the same? (Keep in mind that the volumes you recorded were rounded, not exact values.)
Mathematics
1 answer:
kaheart [24]3 years ago
6 0

Vcylinder : Vcone = 3:1

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Find the coordinates of the vertices of the figure formed by each system of inequalities.
siniylev [52]

Answer:

D(1, –5), (7, 1), (–11, 7)

Step-by-step explanation:

The given inequalities are:

y + x ≥ –4

y ≥ x – 6

3y + x ≤ 10

We graph the inequalities as shown in the attachment.

The coordinates of the vertices of the figure formed by the given system of inequalities are:

(1, –5), (7, 1), (–11, 7)

The correct choice is D.

6 0
3 years ago
A movie was 3 1/2 hours long if Lilly turned of the movie off 1/4 of the way through how much time did she spend watching
Nuetrik [128]

Answer:

7/8 of an hour

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Please HELP ME ON THIS MATH WRITTEN RESPONSE QUESTION.. THANK YOU
sergij07 [2.7K]

Answer:

Quadrilateral ABCD is not a square. The product of slopes of its diagonals is not -1.

Step-by-step explanation:

Point A is (-4,6)

Point B is (-12,-12)

Point C is (6,-18)

Point D is (13,-1)

Given that the diagonals of a square are perpendicular to each other;

We know that the product of slopes of two perpendicular lines is -1.

So, slope(m) of AC × slope(m) of BD should be equal to -1.

Slope of AC = (Change in y-axis) ÷ (Change in x-axis) = (-18 - 6) ÷ (6 - -4) = -24/10 = -2.4

Slope of BD = (Change in y-axis) ÷ (Change in x-axis) = (-1 - -12) ÷ (13 - -12) = 11/25 = 0.44

The product of slope of AC and slope of BD = -2.4 × 0.44 = -1.056

Since the product of slope of AC and slope of BD is not -1 hence AC is not perpendicular to BD thus quadrilateral ABCD is not a square.

4 0
3 years ago
Police estimate that​ 25% of drivers drive without their seat belts. If they stop 6 drivers at​ random, find the probability tha
Furkat [3]

Answer:

17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.

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 combinations 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.

Police estimate that​ 25% of drivers drive without their seat belts.

This means that 75% wear their seatbelts, so p = 0.75

If they stop 6 drivers at​ random, find the probability that all of them are wearing their seat belts.

This is P(X = 6).

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

P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.1780

17.80% probability that all of them are wearing their seat belts.

3 0
3 years ago
Read 2 more answers
A student committee is to consist of 2 freshmen, 5 sophomores, 4 juniors, and 3 seniors. If 6 freshmen, 13 sophomores, 8 juniors
mezya [45]
<h3>Answer:  491,891,400</h3>

Delete the commas if necessary.

============================================================

Explanation:

There are 6 freshmen total and we want to pick 2 of them, where order doesn't matter. The reason it doesn't matter is because each seat on the committee is the same. No member outranks any other. If the positions were labeled "president", "vice president", "secretary", etc, then the order would matter.

Plug n = 6 and r = 2 into the nCr combination formula below

n C r = \frac{n!}{r!(n-r)!}\\\\6 C 2 = \frac{6!}{2!*(6-2)!}\\\\6 C 2 = \frac{6!}{2!*4!}\\\\6 C 2 = \frac{6*5*4!}{2!*4!}\\\\ 6 C 2 = \frac{6*5}{2!}\\\\ 6 C 2 = \frac{6*5}{2*1}\\\\ 6 C 2 = \frac{30}{2}\\\\ 6 C 2 = 15\\\\

This tells us there are 15 ways to pick the 2 freshmen from a pool of 6 total.

Repeat those steps for the other grade levels.

n = 13 sophomores, r = 5 selections leads to nCr = 13C5 = 1287. This is the number of ways to pick the sophomores.

You would follow the same type of steps shown above to get 1287. Let me know if you need to see these steps.

Similarly, 8C4 = 70 is the number of ways to pick the juniors.

Lastly, 14C3 = 364 is the number of ways to pick the seniors.

-----------------------------

To recap, we have...

  • 15 ways to pick the freshmen
  • 1287 ways to pick the sophomores
  • 70 ways to pick the juniors
  • 364 ways to pick the seniors

Multiply out those values to get to the final answer.

15*1287*70*364 = 491,891,400

This massive number is a little under 492 million.

7 0
2 years ago
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