Answer:
Step-by-step explanation:
given that certain tubes manufactured by a company have a mean lifetime of 800 hours and a standard deviation of 60 hours.
Sample size n =16
Std error of sample mean = ![\frac{\sigma}{\sqrt{n} } \\= 15](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%5C%5C%3D%2015)
x bar follows N(800, 15)
the probability that a random sample of 16 tubes taken from the group will have a mean lifetime
(a) between 790 and 810 hours,
=![P(790](https://tex.z-dn.net/?f=P%28790%3Cx%3C810%29%5C%5C%3D%20P%28%7Cz%7C%3C0.667%29%5C%5C%3D%202%2A0.248%5C%5C%3D%200.496)
(b) less than 785 hours
![=P(X](https://tex.z-dn.net/?f=%3DP%28X%3C785%29%5C%5C%3DP%28Z%3C-1%29%5C%5C%3D%200.1584)
, (c) more than 820 hours,
![=P(X>820)\\=p(Z>1.333)\\= 0.0913](https://tex.z-dn.net/?f=%3DP%28X%3E820%29%5C%5C%3Dp%28Z%3E1.333%29%5C%5C%3D%200.0913)
(d) between 770 and 830 hours
=![P(|Z|](https://tex.z-dn.net/?f=P%28%7CZ%7C%3C2%29%5C%5C%3D%200.4772%2A2%5C%5C%3D0.9544)
Answer:
ρ_air = 0.15544 kg/m^3
Step-by-step explanation:
Solution:-
- The deflated ball ( no air ) initially weighs:
m1 = 0.615 kg
- The air is pumped into the ball and weight again. The new reading of the ball's weight is:
m2 = 0.624 kg
- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.
m_air = Δm = m2 - m1
m_air = 0.624 - 0.615
m_air = 0.009 kg
- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:
V_air = 4*π*r^3 / 3
V_air = 4*π*0.24^3 / 3
V_air = 0.05790 m^3
- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:
ρ_air = m_air / V_air
ρ_air = 0.009 / 0.05790
Answer: ρ_air = 0.15544 kg/m^3
Answer:
80
Step-by-step explanation:
You plug in 1200 for x.
C(1200)=0.05(1200) +20
Then solve. Multiply first then add.
60+20
80
Step-by-step explanation:
since it is in the same line as DE, both have the same slope.
the slope is the ratio of "y coordinate change / x coordinate change" when going from one point on the line to another.
and so by using the coordinates of D and E
x changes by +4 (from 8 to 12).
y changes by -8 (from 32 to 24).
the slope is therefore
-8/+4 = -2
Answer:
The total percent decrease is ![$a+b-\frac{a b}{100}$](https://tex.z-dn.net/?f=%24a%2Bb-%5Cfrac%7Ba%20b%7D%7B100%7D%24)
Step-by-step explanation:
- We are given with a word problem
- We are asked to find the total percentage decrease
- We can do this in two steps
Step 1: Finding the two percentage decrease
Step 2: Finding their difference
Step 1 of 2
Let the decrease by a% be 100-a
Then b% decrease is
![$$\begin{aligned}\frac{b}{100} \times(100-a) &=\frac{100 b-a b}{100} \\\frac{100 b}{100}-\frac{a b}{100} &=b-\frac{a b}{100}\end{aligned}$$](https://tex.z-dn.net/?f=%24%24%5Cbegin%7Baligned%7D%5Cfrac%7Bb%7D%7B100%7D%20%5Ctimes%28100-a%29%20%26%3D%5Cfrac%7B100%20b-a%20b%7D%7B100%7D%20%5C%5C%5Cfrac%7B100%20b%7D%7B100%7D-%5Cfrac%7Ba%20b%7D%7B100%7D%20%26%3Db-%5Cfrac%7Ba%20b%7D%7B100%7D%5Cend%7Baligned%7D%24%24)
Step 2 of 2
The percentage difference between the two percentages is,
![$$\begin{aligned}&(100-a)-\left(b-\frac{a b}{100}\right)=100-a-b+\frac{a b}{100} \\&100-a-b+\frac{a b}{100}=100-\left(a+b-\frac{a b}{100}\right)\end{aligned}$$](https://tex.z-dn.net/?f=%24%24%5Cbegin%7Baligned%7D%26%28100-a%29-%5Cleft%28b-%5Cfrac%7Ba%20b%7D%7B100%7D%5Cright%29%3D100-a-b%2B%5Cfrac%7Ba%20b%7D%7B100%7D%20%5C%5C%26100-a-b%2B%5Cfrac%7Ba%20b%7D%7B100%7D%3D100-%5Cleft%28a%2Bb-%5Cfrac%7Ba%20b%7D%7B100%7D%5Cright%29%5Cend%7Baligned%7D%24%24)
Since 100 is the total percent
is the percentage decrease.