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NISA [10]
3 years ago
12

Question 2

Mathematics
1 answer:
Rainbow [258]3 years ago
3 0
Answer D is correct
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A round-robin tournament has 66 pairings. How many teams are in the tournament?
elixir [45]
11 teams because \frac{x(x+1)}{2}=66 where we find x to be 11.
6 0
3 years ago
Read 2 more answers
Q=P(R+R) (SOLVE FOR P)
forsale [732]
Isolate the variable by dividing each side by factors that don't contain the variable. p=q/2r
4 0
3 years ago
Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b
Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b

Step-by-step explanation:

We are given the following in the question:

p(x) = 0.1 e^{-0.1x}

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%

b) P(calls last 4 minutes or more)

=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%

c) cumulative distribution function

P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b

6 0
4 years ago
Please help asap i am timed will give brainliest
wariber [46]

Answer

Step-by-step explanation:

4 0
3 years ago
What is the surface area of the rectangular pyramid
posledela

Answer:

<h2>41.74 m²</h2>

Step-by-step explanation:

We have:

rectangle 4.8 m × 3.8 m

two triangles with base b = 4.8 m and height h = 2.6 m

two triangle with base b = 3.8 m and height h = 2.9 m.

The formula of an area of a rectangle l × w:

A = lw

Substitute:

A_1 = (4.8)(3.8) = 18.24\ m^2

The formula of an area of a triangle:

A=\dfrac{bh}{2}

Substitute:

A_2=\dfrac{(4.8)(2.6)}{2}=6.24\ m^2\\\\A_3=\dfrac{(3.8)(2.9)}{2}=5.51\ m^2

The Surface Area:

S.A.=A_1+2A_2+2A_3\\\\S.A.=18.24+2(6.24)+2(5.51)=41.74\ m^2

3 0
3 years ago
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