Answer:
the answer is B
Step-by-step explanation:
Your welcome.
Answer:
a) Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
b) If the true mean is 190 days, Type II error can be made.
Step-by-step explanation:
Let mu be the mean life of the batteries of the company when it is used in a wireless mouse
Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
Type II error happens if we fail to reject the null hypothesis, when actually the alternative hypothesis is true.
That is if we conclude that mean life of the batteries of the company when it is used in a wireless mouse is at most 183 days, but actually mean life is 190 hours, we make a Type II error.
Option C is your answer.
a² + b² = c²
4² + s² = (√32)²
16 + s² = 32
s² = 16
s = 4
Answer:
See below
Step-by-step explanation:
a) <u>Using the first two lines to get the equation:</u>
Since t = 0 represents a start point, the y-intercept is 163488
<u>Slope is:</u>
- (168392 - 163488)/10 = 490.4
<u>And the equation:</u>
- P(t) = 490.4(t - 1970) + 163488
b) Prediction of the population in 2012 using the function:
- P(2012) = 490.4(2012 - 1970) + 163488 = 184084.8
As we see the number we got is less than the one on the line 3 of the table. So the model underestimated the actual population.
Answer:
(a) 0.06154
(b) 0.2389
(c) 0.6052
(d) 2478
Step-by-step explanation:
probability density function of the time to failure of an electronic component in a copier (in hours) is
P(x) = 1/1076e^−x/1076
λ = 1/1076
A) A component lasts more than 3000 hours before failure:
P(x>3000) = 1 − e^−3000/1076
= 0.06154
B) A component fails in the interval from 1000 to 2000 hours:
P(1000>x>2000) =1 − e^−2000/1076 − 1 +e^−1000/1076 = e^−1000/1076 − e^−2000/1076 = 0.3948 − 0.1559
= 0.2389
C) A component fails before 1000 hours:
P(x<1000) = 0.6052
D) The number of hours at which 10% of all components have failed:
e^−x/1076 = 0.1
= −x/1076
= ln(0.1)
x =(2.3026)×(1076)
x = 2478
