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vovangra [49]
2 years ago
11

Veata wants to know what percent of students at her school are left-hand dominant. Veata recalls that when students were randoml

y assigned to groups she was the only student in the group of 8 that was left-hand dominant. She also notices that 7 of the 28 students in her art class are left-hand dominant. Which sample should be used to determine the best estimate for the percent of students at the school who are left-hand dominant? Circle one of the following.
randomly assigned group

the entire art class

Explain your reasoning and any concerns about the choices above.
Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
7 0
Randomly assigned group. the art class isn't random and only consists of a select type of people--it is not a representative sample; maybe more left handed people enjoy art than do right handed people! therefore, even though the randomly assigned group is smaller, it is the better estimate.
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a baseball team played 154 games last season the had 35 fewer losses than wins how many games did the team win and how many did
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do 154 / 2 and then add 35 and that is how much they won. do 154 / 2 and then subtract 35 and that is how much they loss

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3 years ago
What is the sum of 10+20?
julsineya [31]

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Step-by-step explanation:

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3 years ago
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
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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
identify an equation in point slope form for the line perpedicular to y=1/4x-7 that passes through (-2,-6)​
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Answer:

perp. -4

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6 0
2 years ago
Can someone help me plzz
Rashid [163]
Let's go through each answer choice and eliminate the choices.

a) 6(2/3) = 4, this is less than 6, making it our correct answer, but still go and check each answer
b) 6(2/3) again = 4, this is less than 6, making this answer choice wrong.
c) 6(3/2) = 9, this is greater than 6, making this answer choice wrong.
d) 6(3/3) = 6, this is equal to 6, making this answer choice wrong.
8 0
3 years ago
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