Answer:
The probability that a light bulb of that brand lasts between 1175 hr and 1610 hr is 0.8524.
Step-by-step explanation:
Given : Suppose a brand of light bulbs is normally distributed, with a mean life of 1400 hr and a standard deviation of 150 hr.
To find : The probability that a light bulb of that brand lasts between 1175 hr and 1610 hr ?
Solution :
Applying z-score formula,
where, 
is population mean
is standard deviation
For x=1175 hour,
For x=1610 hour,
The required probability is,
Using z table, the values are
The probability that a light bulb of that brand lasts between 1175 hr and 1610 hr is 0.8524.
Answer:
92.18 feet
Step-by-step explanation:
t(n)=125(.97)^x
t(10)=124(.97)^10
Sequence cant have partial or discrete bounce
Answer:
0
Step-by-step explanation:
The power is 0 since 2^0 = 1.
Answer:
180.
Step-by-step explanation:
Given
x = 1%2F2y
x+y=180
...
Substitute x into the second equation and solve for y, the larger angle
1%2F2y + y = 180
Multiply both sides by 2 to be rid of the fraction
y + 2y = 360
3y = 360
highlight_green%28y=120%29
...
Check cartoon%28%0D%0A120+%2B+%281%2F2%29%28120%29%2C%0D%0A120+%2B+60%2C%0D%0A180%29
Length of one side of the rhombus is 10.
If there is a rhombus EFGH, such as yours, with points E, F, G, H (such as in the .jpeg image in attachment), and if the notation of rhombus side is a (rhombus has 4 sides and each of them is of the same length):
FH/2 = 12/2 = 6 (shorter diagonal of rhombus, it's half)
EG/2 = 16/2 = 8 (longer diagonal of rhombus, it's half)
-----------------------------------------
a^2 = (FH/2)^2 + (EG/2)^2
a^2 = 6^2 + 8^2
a^2 = 36 + 64
a^2 = 100
a = square root of 100
a = 10
