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ivann1987 [24]
3 years ago
15

A gymnasium can hold no more than 650 people. A permanent permanent bleacher in the gymnasium holds 136 people. The event organi

zers are setting up 25 rows with an equal number of chairs. At most, how many chairs can be in each room?
Mathematics
2 answers:
sattari [20]3 years ago
8 0
136 + 25x ≤ 650 
<span>136 - 136 + 25x ≤ 650 - 136 </span>
<span>25x ≤ 514 </span>
<span>(1/25)(25x) ≤ (514)(1/25) </span>
<span>x ≤ 20.56 </span>
<span>there can be at at the most, 20 chair in each row.</span>
Kipish [7]3 years ago
7 0

Answer:

Total number of chairs in each room = 636 chairs

Step-by-step explanation:

Given

Maximum people = 650

Permanent Bleacher holds 136 people

Suppose there are 25 rows set up of seats; let n = number of chairs per row.

This means that there are (650 - 136) seat spaces left

650 - 136 = 514 seat spaces

This means that the total set up of seat must be less than or equal to 514.

Mathematically,

25 * n <= 514 ---Divide through by 25

25n/25 <= 514/25

n <= 20.56

Since n <= 20.56 then number of rows = 20 (Approximated)

Total number of chairs in each room = 20 * 25 + 136

Total number of chairs in each room = 500 + 136

<em>Total number of chairs in each room = 636 chairs</em>

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3 years ago
When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let
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Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = <em>the number of defective boards in a random sample of size, n = 25</em>

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r}  ; x = 0,1,2,......

where, n = number of trials (samples) taken = 25

            r = number of success

            p = probability of success which in our question is percentage

                   of defectivs, i.e. 5%

(a) P(X = 3) =  \binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

                   =  2300 \times 0.05^{3}\times 0.95^{22}

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(b) P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= \binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

=  1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}

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                    =  1 - 0.966

                    =  <u>0.034</u>

<u></u>

(d) P(1 ≤ X ≤ 3) =  P(X = 1) + P(X = 2) + P(X = 3)

=  \binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

=  25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}

=  <u>0.688</u>

(e) The probability that none of the 25 boards is defective is given by = P(X = 0)

     P(X = 0) =  \binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}

                   =  1 \times 1\times 0.95^{25}

                   =  <u>0.277</u>

(f) The expected value of X is given by;

       E(X)  =  n \times p

                =  25 \times 0.05  = 1.25

The standard deviation of X is given by;

        S.D.(X)  =  \sqrt{n \times p \times (1-p)}

                     =  \sqrt{25 \times 0.05 \times (1-0.05)}

                     =  <u>1.089</u>

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