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emmasim [6.3K]
3 years ago
5

What is the area of a parallelogram with a height of 4 ft and length of 8 ft

Mathematics
1 answer:
inna [77]3 years ago
6 0
To find the area of a parallelogram, you simply have to multiply the base and the height. The formula is "A = bh." So, multiply 8 and 4, and you'll get 32. The area of this parallelogram is 32ft.
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Sarah wants to print copies of her artwork. At the local print shop, it costs her $1 to make 5 copies and $5 to make 25 copies.
Ierofanga [76]

Answer:

i wanna say B

Step-by-step explanation:

6 0
3 years ago
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A standard deck of 52 cards contains 13 cards with hearts, 13 with diamonds, 13 with clubs, and 13 with spades. How
PIT_PIT [208]

Answer:

133,784,560

Step-by-step explanation:

using the combinations formula: n!/(n-r!)r!, the equation would be 52!/52-7! x 7! which is simplified to 52!/45!7! which is equal to 133,784,560

7 0
3 years ago
Two friends share 7 cookies how many cookies do each of them get
Goshia [24]

Answer:

3.5 cookies

Step-by-step explanation:

Hi there!

Well, assuming they get the same amount of cookies, you'd do 7/2..

7/2=3.5

So they'd each get 3.5 cookies.

Have a great day! ^u^

7 0
3 years ago
Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin w
Goshia [24]

Answer:

Step-by-step explanation:

Given that;

the following procedure for accomplishing our task are:

1. Flip the coin.

2. Flip the coin again.

From here will know that the coin is first flipped twice

3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails

But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.

Let the probability of head = p

so P(head) = p

the probability of tail be = (1 - p)

This kind of probability follows a conditional distribution and the probability  of getting heads is :

P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})

= \dfrac{P( \{Tails, Heads\})  \cap \{Tails, Heads,( Heads ,Tails)\})}{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) }{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) } {  {P( Tails, Heads) +P( Heads ,Tails)}}

=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}

=\dfrac{(1-p)*p}{2(1-p)*p}

=\dfrac{1}{2}

Thus; the probability of getting heads is \dfrac{1}{2} which typically implies that the coin is fair

(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.

The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.

Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of  p.

In essence, NO, procedure (b) does not give a fair coin flip outcome.

5 0
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Giving braileiest lollolololol
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Answer:

did you try googling it? it works for me

Step-by-step explanation:

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