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Lunna [17]
3 years ago
9

7. Match the correct answers. ​

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
3 0

let's start with the triangle, since that seems very obvious per se, the triangle has a base of 20 and a height of 14.

\textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b= base\\ h= height\\[-0.5em] \hrulefill\\ b=20\\ h=14 \end{cases}\implies A=\cfrac{1}{2}(20)(14)\implies A=140

now let's do the sector.

\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=20\\ \theta =45 \end{cases}\implies \begin{array}{llll} A=\cfrac{(45)\pi (20)^2}{360} \\\\\\ A=50\pi \implies A\approx 157.08 \end{array}

now let's do the segment.

\textit{area of a segment of a circle}\\\\ A=\cfrac{r^2}{2}\left( \cfrac{\pi \theta }{180}~~ - ~~sin(\theta ) \right)~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=20\\ \theta =45 \end{cases} \\\\\\ A=\cfrac{20^2}{2}\left( \cfrac{\pi (45)}{180}~~ - ~~sin(45^o) \right)\implies A=200\left(\cfrac{\pi }{4}~~ - ~~\cfrac{1}{\sqrt{2}} \right) \\\\\\ A=50\pi - 100\sqrt{2}\implies A\approx 15.66

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Answer:

the probability that the user is fraudulent is 0.00299133

Step-by-step explanation:

Let be the events be:

G: The user generates calls from two or more areas.

NG: The user does NOT generate calls from two or more areas.

L: The user is legitimate.

F: The user is fraudulent.

The probabilities established in the statement are:

P (G | L) = 0.01//P (G | F) = 0.30//P (F) = 0.0001//P (L) = 0.9999//

With these values, the probability that a user is fraudulent, if it has originated calls from two or more areas is:

P (F|G) = \frac{P(F\bigcap G)}{P(G)} = \frac{P(F)P(G|F)}{P(G)} = \frac{P(F)P(G|F)}{P(F)P(G|F)+P(L)P(G|L)}

\frac{(0.0001)(0.30)}{(0.0001)(0.30)+(0.9999)(0.01)} = 0.00299133

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Answer:

The maximum distance traveled is 4.73 meters in 0.23 seconds.

Step-by-step explanation:

We have that the distance traveled with respect to time is given by the function,

d(t) = 4.9t^2-2.3t+5.

Now, differentiating this function with respect to time 't', we get,

d'(t)=9.8t-2.3

Equating d'(t) by 0 gives,

9.8t - 2.3 = 0

i.e. 9.8t = 2.3

i.e. t = 0.23 seconds

Substitute this value in d'(t) gives,

d'(t) = 9.8 × 0.23 - 2.3

d'(t) = 2.254 - 2.3

d'(t) = -0.046.

As, d'(t) < 0, we get that the function has the maximum value at t = 0.23 seconds.

Thus, the maximum distance the skateboard can travel is given by,

d(t) = 4.9\times 0.23^2-2.3\times 0.23+5.

i.e. d(t) = 4.9\times 0.0529-2.3\times 0.23+5.

i.e. d(t) = 0.25921-0.529+5.

i.e. d(t) = 0.25921-0.529+5.

i.e. d(t) = 4.73021

Hence, the maximum distance traveled is 4.73 meters in 0.23 seconds.

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