Answer:
See bolded below.
Step-by-step explanation:
Take a look at the attachment below. It represents circle x and y, with respect to each of their radii. In each of the options we are given, we would have to translate and dilate the circles, by a fixed scale factor -
Now the first thing one would do is translate the circles so that they share a common center point, or in other words the center of one circle rests on the edge of the other circle. That way when dilating circle y, it may fit into circle x as it expands.
The second point is how much this smaller circle ( circle y ) has to expand. The radius of circle y being 2, has to increase by 3 times the value to equal the radius of circle x, and hence has to dilate by a scale factor of 3 as to match circle x,
<u><em>Solution = " Translate the circles so the center of one circle rests on the edge of the other circle, and dilate circle Y by a scale factor of 3 " / Option D</em></u>
Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
<em>Firstly we will calculate the general probability that of P(a < X < b) </em>
P(a < X < b) = =
= { Because }
= =
= =
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = = = 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/ - 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = = = 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= + (1/ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ = 0.75
⇒ = 1 - 0.75 = 0.25
⇒ = ⇒ = 4 ⇒ x =
Therefore, value of x such that P(X < x) = 0.75 is 2.
Answer:
Quotient of a number and 3 = x/3
Ten less the quotient of a number and 3 will equal 6
So...
10 - x/3 = 6
+ x/3 +x/3
10 = x/3 +6
-6 - 6
3 × 4=(x/3) × 3
12 = x
1. There are 6 green bulbs because
13+5=18*1/3=6 green bulbs
2. The probability of getting a red bulb is 15/8
Hope it helps you!!!!
Answer:
B
Step-by-step explanation:
I put the answer in an atachement