Answer:
P ( 1.2 < X < 2.1 ) = 0.3
Step-by-step explanation:
Given:
Uniform distribution over interval (0,3) can be modeled by a probability density function f(x)
f(x) = 1 / (b - a)
Where a < x < b is the domain at which function is defined:
f(x) = 1 / (3) = 1 / 3
Where, X - U ( u , δ )
u = ( a + b ) / 2 = (0 +3) / 2 = 1.5
δ = ( b - a ) / sqrt (12) = (3 - 0) / sqrt (12) = 0.866
Hence,
X - U ( 1.5 , 0.866 )
There-fore calculating P ( 1.2 < X < 2.1 ):
Where, a = 1.2 and b = 2.1
P ( 1.2 < X < 2.1 ) = x / 3 |
P ( 1.2 < X < 2.1 ) = 2.1 /3 - 1.2 / 3 = 0.3
Answer: P ( 1.2 < X < 2.1 ) = 0.3
Answer:
108.90
Step-by-step explanation:
You start with finding
10
%
of
90
.
$
90
⋅
.1
=
$
9
You add the money and the interest to get
$
99
for the first year. However, the next year is different. It's compound interest, so you have to multiply
10
%
by
$
99
.
$
99
⋅
.1
=
$
9.90
$
99
+
$
9.90
=
$ 108.90 The total would then be $
108.90 i think lol
Answer:
a) x+3x/2y or (3x+2xy)/2y
b) b^12
c) 4^3 or 64
Step-by-step explanation:
Answer:
The right answer is:
"The one-day SAT prep class produces statistically significant improvements in SAT writing performance."
Step-by-step explanation:
The P-value gives us the probability of getting the sample results, given that the null hypopthesis is true. That is why, it the P-value is small enough (smaller than the level of significance) is an indicator that the null hypothesis may not be true.
In this case, the P-value (0.028) is smaller than the level of significance (0.05), so the null hypothesis is rejected and we can conclude that there is significant difference for the studentes that take the SAT prep-class.
The population only include the students that take teh SAT prep-class, so it is not useful to compare to the performance of students that didn't take the class.
The right answer is:
"The one-day SAT prep class produces statistically significant improvements in SAT writing performance."
