It is the third number line.
b-11=-8
What minus 11 would equal -8? 3!
8+3=11 so it would make sense that the dot would be on 3.
smaller numbers minus larger numbers (such as this equation) always end up being negative, so if you have the answer and at least one of the other numbers, it will be easy to figure out.
In this case you would want to maybe just imagine 11-8=? or go 8+?=11.
11-8=3 and 8+3=11. There you would have your missing number.
The answer is 2, yous add all of them up and divide by how many there are.
The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118 .
In the question ,
it is given that ,
the probability of committing at least , type I error is = 0.7
we have to find the number of tests ,
let the number of test be n ,
the above mentioned situation can be written as
1 - P(no type I error is committed) ≥ P(at least type I error is committed)
which is written as ,
1 - (1 - 0.01)ⁿ ≥ 0.7
-(0.99)ⁿ ≥ 0.7 - 1
(0.99)ⁿ ≤ 0.3
On further simplification ,
we get ,
n ≈ 118 .
Therefore , the number of tests are 118 .
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Answer:
a+305/36
Step-by-step explanation:
Convert
3
1
4
3
1
4
to an improper fraction.
a
+
13
4
+
5
2
9
a
+
13
4
+
5
2
9
Convert
5
2
9
5
2
9
to an improper fraction.
a
+
13
4
+
47
9
a
+
13
4
+
47
9
To write
13
4
13
4
as a fraction with a common denominator, multiply by
9
9
9
9
.
a
+
13
4
⋅
9
9
+
47
9
a
+
13
4
⋅
9
9
+
47
9
To write
47
9
47
9
as a fraction with a common denominator, multiply by
4
4
4
4
.
a
+
13
4
⋅
9
9
+
47
9
⋅
4
4
a
+
13
4
⋅
9
9
+
47
9
⋅
4
4
Write each expression with a common denominator of
36
36
, by multiplying each by an appropriate factor of
1
1
.
a
+
13
⋅
9
36
+
47
⋅
4
36
a
+
13
⋅
9
36
+
47
⋅
4
36
Combine the numerators over the common denominator.
a
+
13
⋅
9
+
47
⋅
4
36
a
+
13
⋅
9
+
47
⋅
4
36
Simplify the numerator.
a
+
305
36
a
+
305
36
