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Ostrovityanka [42]
3 years ago
15

The Lakewood Wildcats won 5 out of their first seven games in this year. There are 28 games in the season. About how many games

would you expect the Wildcats to win the season?
Mathematics
1 answer:
allochka39001 [22]3 years ago
5 0

Answer:

The total number of matches expected to be won by Lakewood Wildcats in this season  is 20.

Step-by-step explanation:

The number of games played by Lakewood Wildcats = 7

Number of matches won by Lakewood = 5

or, the ratio of Won : Played = 5: 7

Total numbers of games in the season = 28

Let the Lakewood Wildcats win m number of games.

Here, ratio of Won Matches : Played Matches = m : 28

Now, by RATIO OF PROPORTIONALITY:

\frac{5}{7}   = \frac{m}{28}

⇒m = \frac{28 \times 5}{7}  = 20

or, m = 20

Hence, the total number of matches expected to be won by Lakewood Wildcats out of total 28 matches is 20.

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Help please and thank you
azamat

Answer:

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Step-by-step explanation:

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7 0
3 years ago
Assume that 63% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below
mash [69]

Answer:

a. 0.9931

b. 0.3423

c. 0.3907

d. 0.2670

e. 3.15

f. 1.0796

Step-by-step explanation:

The probability of the variable that said the number of left-handed people follow a Binomial distribution, so the probability is:

P(x)=nCx*p^{x}*(1-p)^{n-x}

nCx is calculated as:

nCx=\frac{n!}{x!(n-x)!}

Where x is the number of left-handed people, n is the number of people selected at random and p is the probability that a person is left--handed. So P(x) is:

P(x)=5Cx*0.63^{x}*(1-0.63)^{5-x}

Then the probabilities P(0), P(1), P(2), P(3), P(4) and P(5) are:

P(0)=5C0*0.63^{0}*(1-0.63)^{5-0}=0.0069

P(1)=5C1*0.63^{1}*(1-0.63)^{5-1}=0.0590

P(2)=5C2*0.63^{2}*(1-0.63)^{5-2}=0.2011

P(3)=5C3*0.63^{3}*(1-0.63)^{5-3}=0.3423

P(x)=5C4*0.63^{4}*(1-0.63)^{5-4}=0.2914

P(x)=5C5*0.63^{5}*(1-0.63)^{5-5}=0.0993

Then, the probability P(x≥1) that there are some lefties among the 5 people is:

P(x≥1) = P(1) + P(2) + P(3) + P(4) + P(5)

P(x≥1) = 0.0590 + 0.2011 + 0.3423 + 0.2914 + 0.0993 = 0.9931

The probability P(3) that there are exactly 3 lefties in the group is:

P(3) = 0.3423

The probability P(x≥4) that there are at least 4 lefties in the group is:

P(x≥4) = P(4) + P(5) = 0.2914 + 0.0993 = 0.3907

The probability P(x≤2) that there are no more than 2 lefties in the group is:

P(x≤2) = P(0) + P(1) + P(2) = 0.0069 + 0.0590 + 0.2011 = 0.2670

On the other hand, the expected value E(x) and standard deviation S(x) of the variable that follows a binomial distribution is:

E(x)=np=5(0.63)=3.15\\S(x)=\sqrt{np(1-p)}=\sqrt{5(0.63)(1-0.63)}=1.0796

6 0
3 years ago
Can anybody help with this?
pickupchik [31]
I think it’s the last one
5 0
2 years ago
Write down the answer for Q 7 and 8
astra-53 [7]

The answers to the given addition operations are

7) 6 Hundredths add to 4 tenths add to 6 ones is equal to 6.46

8) 82 Hundredths add to 9 tenths add to 4 tens is equal to 41.72

<h3>Addition operation </h3>

From the question, we are to add the given numbers

7. 6 Hundredths add to 4 tenths add to 6 ones is equal to

6 Hundredths = 0.06

4 tenths = 0.4

6 ones = 6

Thus, we get

0.06 + 0.4 + 6 = 6.46

8. 82 Hundredths add to 9 tenths add to 4 tens is equal to

82 Hundredths = 0.82

9 tenths = 0.9

4 tens = 40

Thus, we get

0.82 + 0.9 + 40 = 41.72

Hence, the answers to the given addition operations are

7) 6 Hundredths add to 4 tenths add to 6 ones is equal to 6.46

8) 82 Hundredths add to 9 tenths add to 4 tens is equal to 41.72

Learn more on Addition operation here: brainly.com/question/4248979

#SPJ1

7 0
1 year ago
at the beginning of school, we had 40 algebra books. We increased our supply by 65%. how many algebra books does the school have
valentina_108 [34]
The answer is 95 i guess
6 0
3 years ago
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