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Fudgin [204]
3 years ago
13

(a) Fill in the tavle below to complete the proability distrubution of X

Mathematics
1 answer:
Aliun [14]3 years ago
6 0

Answer:

2.6

Step-by-step explanation:

Given :

x 1 2 3 4

P(x) 0.3 0.2 ? 0.4

Σp(x) = 1

Hence, 0.3 + 0.2 +? + 0.4 = 1

0.9 +? = 1

? = 1 - 0.9

? = 0.1

For the mean :

E(X) = ΣX*P(X)

(1 * 0.3) + (2 * 0.2) + (3 *0.1) + (4* 0.4)

0.3 + 0.4 + 0.3 + 1.6 = 2.6

Hence, the mean = 2.6

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3 years ago
A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,
algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}

Where v_i is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is [0, \pi/2]. The critical  points of the function are those who make d'(\theta)=0:

d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}

d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...

The critical value inside the interval is \pi/4.

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft

The second step is to find the values of the function at the endpoints of the interval:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft

The biggest value of f is gived by \pi/4, therefore \pi/4 is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

4 0
3 years ago
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