Times the whole equation by 15 since the LCM of 3 and 5 is 15.
You get: 5p-3p-6= 60
Rearrange to get: 2p=66
p=33
The sample std. dev. will be (14 inches) / sqrt(49), or (14 inches) / 7, or 2 inches.
Find the z score for 93.8 inches:
93.8 inches - 91.0 inches 2.8 inches
z = ------------------------------------- = ----------------- = 1.4
2 inches 2 inches
Now find the area under the standard normal curve to the left of z = +1.4.
My calculator returns the following:
normalcdf(-100,1.4) = 0.919. This is the probability that the mean annual precipitation during those 49 years will be less than 93.8 inches.
Answer: 286 minutes
Step-by-step explanation:
x : # of months that has gone by
0.24x : cost of the 24 cent plan after "x" minutes
39.95 + 0.10x : cost of the 10 cent plan after "x" minutes
1. 39.95 + 0.10x > 0.24x
2. 39.95 > 0.24x - 0.10x
3. 39.95 > 0.14x
4. 285.36 > x
x must be AT LEAST 286 minutes for plan #2 (39.95 + 0.10x) to be a better deal
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:
Step-by-step explanation:
-5 + (8 x -7)
using PEMDAS, (8 x -7) = -56
so -5 + -56 = -61