[0,1] could not be used to represent a probability.
The inequalities which matches the graph are: x ≥ ₋1.5 and ₋1.5 ≤ x
Given, a number line is moving from ₋3 to ₊5 .
Next a mark is made at ₋1.5 and everything to its left is shaded which means not visible.
When we mark the point and shade the left part of it then we can start applying the inequality expressions.
And from that we can match the applicable inequalities while observing the graph.
- For the first inequality ₋1.5 ≥ x.Here,x value ranges from ₋1.5 to ₊5, hence we take this as an inequality expression.
- Next, if we consider x ≤ ₋1.5, then here x value will range from ₋1.5 to ₋3. where the region is shaded. Hence this expression doesn't satisfy the graph.
- the next expression is ₋1.5 ≤ x. here the value will again range in the shaded area so it is not applicable.
- ₋1.5 ≥ x, here the values will satisfy the graph.
- remaining inequality expressions does not support the graph.
Therefore the only inequalities the graph represents is x ≥ ₋1.5 and ₋1.5 ≤ x
Learn more about "Linear Inequalities" here-
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Answer:
The expected value of the safe bet equal $0
Step-by-step explanation:
If
is a finite numeric sample space and
for k=1, 2,..., n
is its probability distribution, then the expected value of the distribution is defined as
What is the expected value of the safe bet?
In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is
S = {100,-100}
Since the coin is supposed to be fair,
P(X=100)=0.5
P(X=-100)=0.5
and the expected value is
E(X) = 100*0.5 - 100*0.5 = 0
All three numbers multiplied to get the volume
surface area is the formula times base 1 and base 2 times base 3 and answer
