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RoseWind [281]
3 years ago
13

45% of the 647 students at Royal Oak High attended the football game. How many students did not attend?

Mathematics
2 answers:
tresset_1 [31]3 years ago
8 0

Answer:

647 - 0.45(647); 0.55(647)

If 45% represents the number of students that attended, 55% represents the number that did not attend. Change decimals to per cent.

Find .45 of 647 students and subtract.

OR

Just multiply .55 times the total number of students.

Step-by-step explanation:

Lostsunrise [7]3 years ago
3 0
First multiply
.45*647= 291.15
291.15 is the number of people that attended 
subtract 291.15 from 647 and you get 355.85 wait.. but that's not possible so just round it so the answer to your question is 355 people did not attened
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Sindrei [870]
Question 1

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Question 2

600 pretzels/10 minutes
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700 people/ 2 tons
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3 years ago
Can any decimal that ends with a digit in the thousandths place can be written as a fraction with a denominator that is divisibl
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A geologist has collected 5 specimens of basaltic rock and 7 specimens of granite. The geologist instructs a laboratory assistan
MaRussiya [10]

The rocks are chosen without replacement, which means that the hypergeometric distribution is used to solve this question. First we get the parameters, and then we answer the questions. From this, we get that:

  • E(X) = 5.25, Var(X) = 0.5966
  • P(X < 6) = 0.9545
  • P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

The mean and the variance are:

\mu = \frac{nk}{N}

\sigma^2 = \frac{nk(N-k)(N-n)}{N^2(N-1)}

We have that:

5 + 7 = 12 rocks, which means that N = 12

9 are chosen, which means that n = 9

7 are granite, which means that k = 7

Question a:

E(X) = \mu = \frac{9\times7}{12} = 5.25

Var(X) = \sigma^2 = \frac{9\times7(12-7)(12-9)}{12^2(12-1)} = 0.5966

Thus:

E(X) = 5.25, Var(X) = 0.5966

Question b:

Since there are only 5 specimens of basaltic rock, at least 9 - 5 = 4 specimens of granite are needed, which means that:

P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6)

In which

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 4) = h(4,12,9,7) = \frac{C_{7,4}*C_{5,5}}{C_{12,9}} = 0.1591

P(X = 5) = h(5,12,9,7) = \frac{C_{7,5}*C_{5,4}}{C_{12,9}} = 0.4773

P(X = 6) = h(6,12,9,7) = \frac{C_{7,6}*C_{5,3}}{C_{12,9}} = 0.3181

Thus

P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.1591 + 0.4773 + 0.3181 = 0.9545

So P(X < 6) = 0.9545.

Question c:

5 of basaltic and 4 of granite: 0.1591 probability.

7 of granite is P(X = 7), in which

P(X = 7) = h(7,12,9,7) = \frac{C_{7,7}*C_{5,2}}{C_{12,9}} = 0.0455

0.1591 + 0.0455 = 0.2046, thus:

P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

A similar question is found at brainly.com/question/24008577

5 0
3 years ago
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