The expected value of this policy to the insurance company is $285.00.
Using this formula
Policy expected value=Insurance policy charges-[(Probability × Claim)+(Probability × Claim)]
Let plug in the formula
Policy expected value=$1,300-{(.0041)($150,000)+(.08)($5,000)]
Policy expected value=$1,300-($615+$$400)
Policy expected value=$1,300-$1,015
Policy expected value=$285.00
Inconclusion the expected value of this policy to the insurance company is $285.00
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Answer:
There is sufficient evidence that fuel economy goal has been attained.
Step-by-step explanation:
The hypothesis :
H0 : μ < 30.2
H1 : μ ≥ 30.2
The test statistic :
(xbar - μ) ÷ (s/√(n))
xbar = 32.12 ; s = 4.83 ; n = 50
Test statistic :
(32.12 - 30.2) ÷ (4.83/√(50))
1.92 ÷ 0.6830651
T = 2.811
Using the Pvalue from test statistic calculator :
Since we used the sample standard deviation, we use the T distribution
df = n - 1 = 50 - 1 = 49
Pvalue(2.811, 49) ; one tailed = 0.00354
At α = 0.05
Pvalue < α ; then we reject the null and conclude that there is sufficient evidence that fuel economy goal has been attained
6a+8b; when a variable is next to a coefficient then that means they are multiplying each other, we just don't know what a or b is.
It would be 12:00 in London if its 13:00/1:00 in South Africa
Answer:
using a negative facto to factor the expression will be:
- 7a+49b - 35 = -7(a-7b+5)
Step-by-step explanation:
Given the expression
- 7a+49b - 35
Rewrite -35 as 5 · 7
Rewrite 49 as 7 · 7
so
= -7a + 7 · 7b - 5 · 7
as -7 is the common factor, so factoring out
= -7(a-7b+5)
Thus, using a negative facto to factor the expression will be:
- 7a+49b - 35 = -7(a-7b+5)
