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Delvig [45]
3 years ago
7

Please help me it's the lateral area of a cone along with the surface area

Mathematics
1 answer:
jekas [21]3 years ago
3 0

Answer:

the lateral area of cone is L=A﹣πr2

the surface area of cone is

a = \pi \: r \sqrt{h^{2} + r {}^{2}  }

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Helppp meeee pleaseeee
Alex_Xolod [135]

Answer:

y ≥ -2

Step-by-step explanation:

5y - 3 ≥ -13

   +3     +3

5y ≥ -10

5/5y ≥ -10/5

y ≥ -2

and since we didn't divide by any negative numbers the sign stays the same

6 0
3 years ago
If one class collected 65 cans for the food drive and another class collected 176 cans, how many cans did the two classes collec
kotegsom [21]
In this equation C means the total. In order to find the total you must add the two class together to get the total. The equation that represents this would be the fourth one. 65 + 176 = C
7 0
4 years ago
Read 2 more answers
if 8 men can build a house in 90 days, in how many days can 20 men working under the same rate build the house?​
ikadub [295]

It will take 36 days for 20 men to complete the building

4 0
3 years ago
Bob has a standard deck of playing cards. If he randomly draws a J, K, 2, and 2, what is the probability that the next card he d
nlexa [21]

Answer:

The correct answer has already been given (twice). I'd like to present two solutions that expand on (and explain more completely) the reasoning of the ones already given.


One is using the hypergeometric distribution, which is meant exactly for the type of problem you describe (sampling without replacement):


P(X=k)=(Kk)(N−Kn−k)(Nn)


where N is the total number of cards in the deck, K is the total number of ace cards in the deck, k is the number of ace cards you intend to select, and n is the number of cards overall that you intend to select.


P(X=2)=(42)(480)(522)


P(X=2)=61326=1221


In essence, this would give you the number of possible combinations of drawing two of the four ace cards in the deck (6, already enumerated by Ravish) over the number of possible combinations of drawing any two cards out of the 52 in the deck (1326). This is the way Ravish chose to solve the problem.


Another way is using simple probabilities and combinations:


P(X=2)=(4C1∗152)∗(3C1∗151)


P(X=2)=452∗351=1221


The chance of picking an ace for the first time (same as the chance of picking any card for the first time) is 1/52, multiplied by the number of ways you can pick one of the four aces in the deck, 4C1. This probability is multiplied by the probability of picking a card for the second time (1/51) times the number of ways to get one of the three remaining aces (3C1). This is the way Larry chose to solve the this.

Step-by-step explanation:


6 0
3 years ago
Read 2 more answers
In a race between 18 people, how many ways can the top 5 finishers be arranged?
snow_lady [41]

Answer:

Step-by-step explanation:

This is permutation, since order matters.  The formula for us is

₁₈P₅ = \frac{18!}{(18-5)!} which simplifies to

₁₈P₅ = \frac{18*17*16*15*14*13!}{13!}

The factorial of 13 cancels out on the top and bottom leaving you with

₁₈P₅ = 18 × 17 × 16 × 15 × 14

which comes to 1,028,160 ways

Another way to look at it is:  the first 5 people of 18 finish and the others you don't care about.  Once the first place person is first, there are only 4 of the 18 left to finish in second place.  Then there are only 3 left to finish in third place, etc.  So if we use that reasoning, we don't even need to use the formula, we can just say

18 * 17 * 16 * 15 * 14 and those are the first 5 people of 18 to finish.

8 0
4 years ago
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