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2 years ago

PLEASE HELP with proving lines parallel

Mathematics

1 answer:

den301095 [7]2 years ago

4 0

90 mins later we had to go to sleep for sub for two other people that were not

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write a system of equations for the problem, and then solve the system. if a plane can travel 340 miles per hour with the wind o

BlackZzzverrR [31]

Answer:

w = 40 miles per hour

r = 300 miles per hour

Step-by-step explanation:

We add w with the wind and subtract w against the wind

r+w = 340 miles per hour

r - w = 260 miles per hour

Add the two equations together to eliminate w

r+w = 340 miles per hour

r - w = 260 miles per hour

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

2r = 600

Divide by 2

r = 300 miles per hour without the wind

Now find w

r+w = 340

300 + w = 340

Subtract 300

w = 340-300

w = 40 miles per hour

7 0

3 years ago

A lacrosse team won 80% of their games. If they won 24 games, how many games did they play?

Phantasy [73]

Answer:

Step-by-step explanation:

80% of 30 is 24 (u can use a calculator if you would like to check the answer but it is 30)

8 0

2 years ago

Read 2 more answers

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

Read 2 more answers

Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi and enter your answer as a decimal

timurjin [86]

Answer:

fefe efwe wefw efwffwee ef ewwfwef we fwefwe we

Step-by-step explanation:

fweef efwgfwe r2323 fewerf wefwef23 ew2fefeff ej ufue f2343 4 343 433 3

7 0

2 years ago

Find the simple interest.

Mila [183]

<span>900(1.03)^5
900*1.1593=1,043.37 answer

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6 0

3 years ago