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3 years ago

the volume of a cone is 1,400 cm3. if the radius of the cone is 4cm. what is the approximate height of the cone?

Mathematics

2 answers:

Mandarinka [93]3 years ago

8 0

Answer:

i dont know sorry

Step-by-step explanation:

katrin2010 [14]3 years ago

5 0

Answer:

Step-by-step explanation:

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Please solve this for 15 points

Travka [436]

Answer:5/6 7/8 because the cm get turned into seconds

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3 years ago

Please assist with this 2 column proof

OverLord2011 [107]

Answer:

use the symmetric and transitive properties of congruence

Step-by-step explanation:

(1) ∠ABC ≅ ∠DEF and ∠GHI ≅ ∠DEF; given

(2) ∠DEF ≅ ∠GHI; symmetric property of congruence

(3) ∠ABC ≅ ∠GHI; transitive property of congruence (QED)

_____

Additional comment

The transitive property says if A≅B and B≅C, then A≅C. We used the symmetric property to swap sides of the congruence symbol with DEF and GHI to put the statement in the form to match the transitive property.

You will note we used the form ...

(statement number) statement; reason

with statement and reason separated by a semicolon.

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3 years ago

Round 726.1851 to the nearest tenth.

IgorC [24]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .1

snow_lady [41]

We assumed in this answer that the question b is, Are the events V and M independent?

Answer:

(a). The probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ . (b). The events V and M are not independent.

Step-by-step explanation:

The key factor to solve these questions is to know that:

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

We already know from the question the following probabilities:

$\\ P(V) = 0.73$

$\\ P(M) = 0.18$

The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So

$\\ P(V \cap M) = 0.03$

The probability that a student has either a Visa card or a MasterCard

We can interpret this probability as $\\ P(V \cup M)$ or the sum of both events; that is, the probability that one event occurs OR the other.

Thus, having all this information, we can conclude that

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

$\\ P(V \cup M) = 0.73 + 0.18 - 0.03$

$\\ P(V \cup M) = 0.88$

Then, the probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ .

Are the events V and M independent?

A way to solve this question is by using the concept of conditional probabilities.

In Probability, two events are independent when we conclude that

$\\ P(A|B) = P(A)$ [1]

The general formula for a conditional probability or the probability that event A given (or assuming) the event B is as follows:

$\\ P(A|B) = \frac{P(A \cap B)}{P(B)}$

If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that

$\\ P(M|V) = \frac{P(M \cap V)}{P(V)}$

$\\ P(M|V) = \frac{0.03}{0.73}$

$\\ P(M|V) \approx 0.041$

If M were independent from V (according to [1]), we have

$\\ P(M|V) = P(M) = 0.18$

Which is different from we obtained previously;

That is,

$\\ P(M|V) \approx 0.041$

So, the events V and M are not independent.

We can conclude the same if we calculate the probability

$\\ P(V|M)$ , as follows:

$\\ P(V|M) = \frac{P(V \cap M)}{P(M)}$

$\\ P(V|M) = \frac{0.03}{0.18}$

$\\ P(V|M) = 0.1666.....\approx 0.17$

Which is different from

$\\ P(V|M) = P(V) = 0.73$

In the case that both events were independent.

Notice that

$\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)$

$\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03$

$\\ 0.03 = 0.03 = 0.03 = 0.03$

3 0

4 years ago

Find the constant of variation for the relation and use it to write and solve the equation.

Ksenya-84 [330]

Answer:

When x = 1 and z = 4, $y=\frac{80}{9}$

Step-by-step explanation:

The variation described in the problem can be written using a constant of proportionality "b" as:

$y=b\,\,x\,\,z^2$

The other piece of information is that when x = 5 and z = 1, then y gives 25/9. So we use this info to find the constant "b":

$y=b\,\,x\,\,z^2\\\frac{25}{9} =b\,\,(5)\,\,(1)^2\\\frac{25}{9} =b\,\,(5)\\b=\frac{5}{9}$

Knowing this constant, we can find the value of y when x=1 and z=4 as:

$y=b\,\,x\,\,z^2\\y=\frac{5}{9} \,\,x\,\,z^2\\y=\frac{5}{9} \,\,(1)\,\,(4)^2\\y=\frac{5*16}{9}\\y=\frac{80}{9}$

3 0

3 years ago