Given that mean=3750 hours and standard deviation is 300:
Then:
<span>a. The probability that a lamp will last for more than 4,000 hours?
P(x>4000)=1-P(x<4000)
but
P(x<4000)=P(z<Z)
where:
z=(x-</span>μ)/σ
z=(4000-3750)/300
z=0.833333
thus
P(x<4000)=P(z<0.8333)=0.7967
thus
P(x>4000)=1-0.7967=0.2033
<span>b.What is the probability that a lamp will last less than 3,000 hours?
P(x<3000)=P(z<Z)
Z=(3000-3750)/300
z=-2.5
thus
P(x<3000)=P(z<-2.5)=0.0062
c. </span><span>.What lifetime should the manufacturer advertise for these lamps in order that only 4% of the lamps will burn out before the advertised lifetime?
the life time will be found as follows:
let the value be x
the value of z corresponding to 0.04 is z=-2.65
thus
using the formula for z-score:
-2.65=(x-3750)/300
solving for x we get:
-750=x-3750
x=-750+3750
x=3000</span>
Answer:
(3, 4 )
Step-by-step explanation:
5x + 3y = 27 → (1)
2x + y = 10 ( subtract 2x from both sides )
y = 10 - 2x → (2)
substitute y = 10 - 2x into (1)
5x + 3(10 - 2x) = 27
5x + 30 - 6x = 27
- x + 30 = 27 ( subtract 30 from both sides )
- x = - 3 ( multiply both sides by - 1 )
x = 3
substitute x = 3 into (2)
y = 10 - 2(3) = 10 - 6 = 4
solution is (3, 4 )
Answer:
you have to be more specific on what your talking about
Step-by-step explanation:
It depend on what kind of dog for treat
