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3 years ago

9

at a particular school with 200 male students, 58 play football, 40 play basketball, and 8 play both. what is the probability th

at a randomly selected male student plays neither sport?

2 answers:

nignag [31]3 years ago

7 0

this helps (no play) = 121/200 = 0.6050

ifti doesn't I tried

Dominik [7]3 years ago

4 0

Answer:

$\frac{11}{20}$

Step-by-step explanation:

Total male students = 200

Students who play football (F)= 58

Students who play basketball (B)= 40

Students who play both (F∩B)= 8

The number of students who play either sport is

$(F\cup B)=F+B-(F\cap B)$

$(F\cup B)=58+40-8=90$

The number of students who play neither sport is

$200-(F\cup B)=200-90=110$

The probability that a randomly selected male student plays neither sport is

$Probability=\dfrac{\text{Number of students who play neither sport}}{\text{Total male students}}$

$Probability=\dfrac{110}{200}$

$Probability=\dfrac{11}{20}$

Hence, the required probability is $\frac{11}{20}$ .

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3 years ago

Suppose a shoe factory produces both low-grade and high-grade shoes. The factory produces at least twice as many low-grade as hi

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Answer:

The factory should produce 166 pairs of high-grade shoes and 364 pairs of low-grade shoes for maximum profit

Step-by-step explanation:

The given parameters for the shoe production are;

The number of low grade shoes the factory produces ≥ 2 × The number of high-grade shoes produced by the factory

The maximum number of shoes the factory can produce = 500 pairs of shoes

The number of high-grade shoes the dealer calls for daily ≥ 100 pairs

The profit made per pair of high-grade shoed = Birr 2.00

The profit made per of low-grade shoes = Birr 1.00

Let 'H', represent the number of high grade shoes the factory produces and let 'L' represent he number of low-grade shoes the factory produces, we have;

L ≥ 2·H...(1)

L + H ≤ 500...(2)

H ≥ 100...(3)

Total profit, P = 2·H + L

From inequalities (1) and (2), we have;

3·H ≤ 500

H ≤ 500/3 ≈ 166

The maximum number of high-grade shoes that can be produced, H ≤ 166

Therefore, for maximum profit, the factory should produce the maximum number of high-grade shoe pairs, H = 166 pairs

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2 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

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The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

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