Answer:
what do you mean your not giving any questions about what you need help on
We are given the height of Joe which is 1.6 meters, the length of his shadow is 2 meters when he stands 3 meters from the base of the floodlight.
First, we have to illustrate the problem. Then we can observe two right triangles formed, one is using Joe and the length of the shadow, the other is the floodlight and the sum of the distance from the base plus the length of the shadow. To determine the height of the floodlight, use ratio and proportion:
1.6 / 2 = x / (2 +3)
where x is the height of the flood light
solve for x, x = 4. Therefore, the height of the floodlight is 4 meters.
Answer:
y=5/6x+8
Step-by-step explanation
This is definitely parallel
P(A|B)<span>P(A intersect B) = 0.2 = P( B intersect A)
</span>A) P(A intersect B) = <span>P(A|B)*P(B)
Replacing the known vallues:
0.2=</span><span>P(A|B)*0.5
Solving for </span><span>P(A|B):
0.2/0.5=</span><span>P(A|B)*0.5/0.5
0.4=</span><span>P(A|B)
</span><span>P(A|B)=0.4
</span>
B) P(B intersect A) = P(B|A)*P(A)
Replacing the known vallues:
0.2=P(B|A)*0.6
Solving for P(B|A):
0.2/0.6=P(B|A)*0.6/0.6
2/6=P(B|A)
1/3=P(B|A)
P(B|A)=1/3
Answer:
Step-by-step explanation:
Method #1
We can draw a <em>right triangle</em> on the graph upon where the points are located and use the Pythagorean Theorem:
* Whenever we talk about distance, we ONLY want the NON-NEGATIVE root.
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Method #2
Or, we can use the Distance Formula:
![\sqrt{[-x_1 + x_2]^{2} + [-y_1 + y_2]^{2}} = D](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E%7B2%7D%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E%7B2%7D%7D%20%3D%20D)
[2, 7] [3, −3]
![\sqrt{[-3 + 2]^{2} + [3 + 7]^{2}} = D](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-3%20%2B%202%5D%5E%7B2%7D%20%2B%20%5B3%20%2B%207%5D%5E%7B2%7D%7D%20%3D%20D)
![\sqrt{[-1]^{2} + 10^{2}} = D](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-1%5D%5E%7B2%7D%20%2B%2010%5E%7B2%7D%7D%20%3D%20D)
** You see? It does not matter which method you choose, as long as you are doing the work correctly.
I am delighted to assist you anytime.
