Refer to the following formula for expected payoff:
1 answer:
Answer:
Step-by-step explanation:
From the given information:
The price reduction = 98% = 0.98
where;
P(rival not matching ) = (100 - 98)% = 2%
P(rival not matching ) = 0.02
The expected payoff = [(0.98 * -800) + (0.02*50000)]
The expected payoff = [( -784+ 1000)]
The expected payoff = 216
(b) Probability of rivals reducing price = 5%
= 5/100
= 0.05
∴
Probability of rivals reducing price = 1 - 0.05 = 0.95
The expected payoff = (0.05 * -6000) + (0.95 *0)
The expected payoff = -300 + 0
The expected payoff = -300
(c) Yes.
Based on answers (a) and (b), the firm should cut the price.
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Answer:
Step-by-step explanation: I hope you understand this better
Answer/Step-by-step explanation:
Given:
C = right angle = 90°
BC = a = ?
AB = c = 12
AC = b = 9
Required:
a, <A, and <B
Solution:
✔️Find a using Pythagorean Theorem:
a = √(c² - b²)
a = √12² - 9²) = √63
a = 7.93725393 = 8 (nearest whole number)
✔️Find A by applying trigonometric ratio:
Thus,
Reference angle = A
Hypotenuse = 12
Adjacent = 9
Therefore,
m<A = 41° (nearest whole number)
✔️Find B:
m<B = 180 - (90 + 41) (sum of interior angles of triangle)
m<B = 49°
Answer:
This data suggest that there is more variability in low-dose weight gains than in control weight gains.
Step-by-step explanation:
Let be the variance for the population of weight gains for rats given a low dose, and
the variance for the population of weight gains for control rats whose diet did not include the insecticide.
We want to test vs
. We have that the sample standard deviation for
female control rats was
g and for
female low-dose rats was
g. So, we have observed the value
which comes from a F distribution with
degrees of freedom (numerator) and
degrees of freedom (denominator).
As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.