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4 years ago

Luke has 25 units of fencing which is the largest rectangle for which he can use the fencing?

1 answer:

8 0

The perimeter of Rectangle should be equal to 25 units as the length of fence is 25 units.
So here is one example,
length of rectangle is 10 units and width is 2.5 units and perimeter is 2×(l+b)
=> 2×(10+2.5)
=> 2×12.5
=> 25 units.

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Jose wants to put a fence around his rectangular garden. His garden measures 31 feet by 33 feet. The garden has a path around it

vlabodo [156]

3 feet width is added to both sides .
so the dimensions become 34 feet and 36feet
hence the fencing length required or the perimeter= 2(l+b)
= 2( 34+36)
= 2(70)
= 140feet fence

4 0

4 years ago

Vivi is a drummer for a band. She burns 756756756 calories while drumming for 333 hours. She burns the same number of calories e

alexandr1967 [171]

burned: 756 calories

hours: 3

same number of calories each hour

-> to find this, we must split the number of calories evenly among the hours

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756/3

= 252 calories per hour

6 0

3 years ago

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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 = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = 0.0083.

3 0

4 years ago

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ANSWER FOR BRAINLIEST:

AlekseyPX

Step-by-step explanation:

I think mode,range and median dpo ako sure

3 0

3 years ago

What multiplication expression can be broken apart into (3x4) + (3x4) using the distributive property

Anna007 [38]

The original expression here might be (4x3) + (4x3). This is because the distributive property simply moves around numbers, but will not change the value of the equation, as in the question we are looking at here.

5 0

3 years ago

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