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maw [93]
3 years ago
10

What is the length of the shorter base of the Little Trapezoid trail?

Mathematics
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:

The length of the shorter base of the little trapezoid trail is 1 mi.

Step-by-step explanation:

Let the shorter base of the large trapezoid is S and the larger base of the large trapezoid is L.

Similarly, assume that the shorter base of the small trapezoid is s and the larger base of the small trapezoid is l.

Since, the trapezoids are similar, so

\frac{L}{l} = \frac{S}{s}

Now, given that S = 2 mi, L = 8 mi and l = 4 mi and we have to find s.

So, s = \frac{S \times l}{L} = \frac{4 \times 2}{8} = 1 mi. (Answer)

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Answer:

a) 0.9

b) Mean = 1.58

Standard Deviation = 0.89

Step-by-step explanation:

We are given the following in the question:

A marketing firm is considering making up to three new hires.

Let X be the variable describing the number of hiring in the company.

Thus, x can take values 0,1 ,2 and 3.

P(x\geq 2) = 50\%= 0.5\\P(x = 0) = 10\% = 0.1\\P(x = 3) = 18\% = 0.18

a) P(firm will make at least one hire)

P(x\geq 2) = P(x=2) + P(x=3)\\0.5 = P(x=2) + 0.18\\ P(x=2) = 0.32

Also,

P(x= 0) +P(x= 1) + P(x= 2) + P(x= 3) = 1\\ 0.1 + P(x= 1) + 0.32 + 0.18 = 1\\ P(x= 1) = 1- (0.1+0.32+0.18) = 0.4

\text{P(firm will make at least one hire)}\\= P(x\geq 1)\\=P(x=1) + P(x=2) + P(x=3)\\ = 0.4 + 0.32 + 0.18 = 0.9

b) expected value and the standard deviation of the number of hires.

E(X) = \displaystyle\sum x_iP(x_i)\\=0(0.1) + 1(0.4) + 2(0.32)+3(0.18) = 1.58

E(x^2) = \displaystyle\sum x_i^2P(x_i)\\=0(0.1) + 1(0.4) + 4(0.32) +9(0.18) = 3.3\\V(x) = E(x^2)-[E(x)]^2 = 3.3-(1.58)^2 = 0.80\\\text{Standard Deviation} = \sqrt{V(x)} = \sqrt{0.8036} = 0.89

7 0
3 years ago
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