Answer:
84:48
Step-by-step explanation:
7:4 = x:48
7*48 = 4x
7*48/4 = x
x = 7*12 = 84
So, the equivalent ratio is 84:48
Which of the following correctly identifies the set of outputs?
{(5,-2), (1, -1), (-2, 2), (2,5)}
{(-2,5), (-1, 1), (2,-2), (5, 2)}
{-2,-1,2,5)
{-2, 1, 2,5)
Answer:
the second answer
Step-by-step explanation:
it goes from x to y from graphing points
a) 12π
b) 36π
c) 24π
d) 144π
To express 75.4 in terms of π, we divide 75.4 by the approximate value of π (3.14159). The result of this operation is roughly 24. Therefore, 75.4 in terms of π is 24π. the correct answer is (c) 24π.
To convert 75.4 into terms of π, you need to divide the number 75.4 by the approximate value of π, which equals to 3.14159. When you divide 75.4 by 3.14159, you get approximately 24. Therefore, 75.4 in terms of π is 24π. So the correct answer is (c) 24π.
To express 75.4 in terms of π, we divide 75.4 by π. So, the answer is 75.4/π. This cannot be simplified further, so the final answer is 75.4/π.
#SPJ2
Answer:
In both parts of the question, you're asked to find the limit as
n approaches infinity for certain probabilities involving the estimation of the unknown probability of success
π. Given that
=
0.5
π=0.5, we can simplify the expressions and apply limit properties.
Step-by-step explanation:
Let's start with part (a):
a) Find
lim
�
→
∞
�
(
�
(
�
^
−
�
)
≤
�
)
lim
n→∞
P(n(
π
^
−π)≤x), if
�
=
0.5
π=0.5.
In a Bernoulli distribution, the variance of the estimator
�
^
π
^
is given by
Var
(
�
^
)
=
�
(
1
−
�
)
�
Var(
π
^
)=
n
π(1−π)
. Since
�
=
0.5
π=0.5, this variance simplifies to
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
.
We can use the Central Limit Theorem (CLT) here. The CLT states that as
�
n approaches infinity, the distribution of the sample mean approaches a normal distribution with mean
�
μ (population mean) and variance
�
2
�
n
σ
2
, where
�
2
σ
2
is the population variance. Since we have
�
=
0.5
π=0.5 and
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
, we can treat
�
^
π
^
as a sample mean of Bernoulli trials with
�
=
0.5
π=0.5.
Now, let's rewrite the expression
lim
�
→
∞
�
(
�
(
�
^
−
�
)
≤
�
)
lim
n→∞
P(n(
π
^
−π)≤x) as a z-score (standard score) and find the limit:
lim
�
→
∞
�
(
�
(
�
^
−
�
)
Var
(
�
^
)
≤
�
Var
(
�
^
)
)
lim
n→∞
P(
Var(
π
^
)
n(
π
^
−π)
≤
Var(
π
^
)
x
)
Substitute the values:
�
=
0.5
π=0.5 and
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
:
lim
�
→
∞
�
(
2
�
(
�
^
−
0.5
)
1
4
�
≤
�
1
4
�
)
lim
n→∞
P(
4n
1
2n(
π
^
−0.5)
≤
4n
1
x
)
Simplify:
lim
�
→
∞
�
(
4
�
(
�
^
−
0.5
)
≤
2
�
)
lim
n→∞
P(4n(
π
^
−0.5)≤2x)
Notice that the left-hand side now resembles a z-score. As
�
n goes to infinity, the expression will converge to the standard normal distribution's cumulative distribution function (CDF). Therefore, the limit is:
lim
�
→
∞
�
(
4
�
(
�
^
−
0.5
)
≤
2
�
)
=
Φ
(
2
�
)
lim
n→∞
P(4n(
π
^
−0.5)≤2x)=Φ(2x)
where
Φ
Φ represents the standard normal cumulative distribution function.
This limit is not dependent on
�
π and will approach the value of
Φ
(
2
�
)
Φ(2x) as
�
n goes to infinity.
For part (b), the approach is similar, but it involves the logit transformation. The logit transformation of
�
^
π
^
is
logit
(
�
^
)
=
log
(
�
^
1
−
�
^
)
logit(
π
^
)=log(
1−
π
^
π
^
). You would follow a similar process of simplifying and finding the limit as
�
n approaches infinity.
1
6
Answer:
16 which is twice as never mind let me say what you said,
Step-by-step explanation: A fair six-sided die is rolled twice. What is the theoretical probability that the first number that comes up is greater than or equal to the second number?
A.
1
6
Answer:
1/6
Step-by-step explanation:
1. this is because a normal dice is 1/6