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alexandr1967 [171]
3 years ago
10

What’s the answer ? Please reply fast !

Mathematics
1 answer:
mezya [45]3 years ago
3 0

Answer:7.5

Step-by-step explanation:

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A local park has a grassy playground that is 3/5 mile long and 1/3 mile wide. What part of a square mile does the playground cov
adell [148]

Answer:

The playground covers \frac{1}{5} of a square mile


Step-by-step explanation:

The playground is a rectangular ground. The area of a rectangle is given by legth * width, hence,

The area of the playground is length * width

Area = \frac{3}{5}*\frac{1}{3}=\frac{3}{15}=\frac{1}{5} sq. mi


Hence the playground cover \frac{1}{5} of a square mile

5 0
3 years ago
Let X be the number of the cars being repaired at a repair shop. We have the following information:
worty [1.4K]

Answer:

(a) Sample Space

S = \{0,1,2,3\}

(b) PMF

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}

(c) CDF

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

S = \{0,1,2,3\}

Solving (b): The PMF

From the question, we have:

P(2) = P(1)

P(0) = P(3)

P(1\ or\ 2) = 0.5 * P(0\ or\ 3)

P(1\ or\ 2) = 0.5 * P(0\ or\ 3) can be represented as:

P(1) + P(2) = 0.5[P(0) + P(3)]

Substitute P(2) = P(1) and P(0) = P(3)

P(1) + P(1) = 0.5[P(0) + P(0)]

2P(1) = 0.5[2P(0)]

2P(1) = P(0)

P(0)= 2P(1)

Also note that:

P(0) + P(1) + P(2) + P(3) = 1

Substitute P(2) = P(1) and P(0) = P(3)

P(0) + P(1) + P(1) + P(0) = 1

2P(1) + 2P(0) = 1

Substitute P(0)= 2P(1)

2P(1) + 2*2P(1) = 1

2P(1) + 4P(1) = 1

6P(1) = 1

Solve for P(1)

P(1) = \frac{1}{6}

To calculate others, we have:

P(2) = P(1)

P(2) = P(1) = \frac{1}{6}

P(0)= 2P(1)

P(0) =2 * \frac{1}{6}P(0) =\frac{1}{3}

P(3) = P(0) =\frac{1}{3}

Hence, the PMF is:

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}

<em>See attachment (1) for histogram</em>

Solving (c): The CDF ; F(x)

This is calculated as:

F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x}  P(x_i)

For x = 0;

We have:

P(X \le 0) = P(0)

P(X \le 0) = 1/3

For x = 1

P(X \le 1) = P(0) + P(1)

P(X \le 1) =  1/3 + 1/6

P(X \le 1) =  1/2

For x = 2

P(X \le 2) =  P(0) + P(1) + P(2)

P(X \le 2) =  1/3 + 1/6 + 1/6

P(X \le 2) =  2/3

For x = 3

P(X \le 3) =  P(0) + P(1) + P(2) + P(3)

P(X \le 3) =  1/3 + 1/6 + 1/6 + 1/3

P(X \le 3) =  1

Hence, the CDF is:

\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}

<em>See attachment (2) for histogram</em>

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Answer:

what do u need to find in here?

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