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:
<h2><u>
=</u>
<u>
57
/ 514 </u>
<u>
(Decimal: 0.110895)</u></h2>
Step-by-step explanation:
57
/ 514
<u>= 57
/ 514
</u>
<u>(Decimal: 0.110895)</u>
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And if that is not what you are looking for here: </u></h2><h2><u>
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Rewrite the equation as
x
/14
= 5/
7
. x/
14
= 5/
7
Multiply both sides of the equation by
14.14 ⋅ x
/14
= 14
⋅
5
/7
Simplify both sides of the equation.
Tap for fewer steps...
Cancel the common factor of 14
.
Cancel the common factor.
14
⋅ x
/14
= 14
⋅
5
/7
Rewrite the expression.
x
=
14
⋅
5
/7
Simplify 14
⋅ 5/
7
.
Cancel the common factor of 7
.
Factor 7 out of 14
.
x
=
7
(
2
)
⋅
5/
7
Cancel the common factor.
x
=
7
⋅ 2
⋅ 5/
7
Rewrite the expression.
x =
2
⋅
5
Multiply 2 by 5
.
<u>x
=
10</u>
Answer:
1 e
2 b
Step-by-step explanation:
120 different 3digit numbers
<h3>
Therefore either
or, 
</h3>
Step-by-step explanation:

here a = 3 ,b = 9 and c= -6



Therefore either
or, 