Complete question is;
The abc battery company claims that their batteries last 100 hours, on average. You decide to conduct a test to see if the company's claim is true. You believe that the mean life may be different from the 100 hours the company claims. you decide to collect data on the average battery life (in hours) of a random sample of n = 20 batteries. some of the information related to the hypothesis test is presented below:
Test of H0: μ = 100 versus H1: μ ≠ 100
Sample mean: 98.5
Std error of mean: 0.777
Assuming the life length of batteries is normally distributed, what is the p-value associated with this test?
Answer:
p-value = 0.00001
Explanation:
We are given;
Null hypothesis; H0: μ = 100
Alternative Hypothesis; H1: μ ≠ 100
Sample mean: x = 98.5
Standard error of mean; s = 0.777
To find the test statistic, we will use the formula;
t = (x - μ)/(s/√n)
t = (98.5 - 100)/(0.777/√20)
t = -1.5/0.1737
t = -8.64
Now, from online p-value from t-score calculator attached, using t = -8.64; DF = n - 1 = 20 - 1 = 19; two tail distribution;significance level of 0.05; we have;
The p-value = 0.00001
Answer:
6.7 x 10^8 mi/hr
Explanation:
Do the math with units ONLY
km/s * miles/ km * s/hr = mi / hr ( ...as the km and s cancel out)
now.... put in the numbers
3 x 10^5 km/s * .62 mi/km * 3600 s / hr = 6.7x 10 ^8 mi/hr
Answer:
Explanation:
Refraction: It is a process in which light ray travel from one medium to another medium.
When a light travel form one medium to another medium then the speed of light changes.
When a light travel from rare to denser medium then the light bend towards normal .
When a light travel form denser to rare medium then the light ray bend away from the normal.
Refractive index:It is a ratio of velocity of light in vacuum to the velocity of light in medium .It is a measure of change in speed of light when it passes from vacuum into material.It is represented by "n''.
Mathematical representation:
Where n=Refractive index of material
c=Velocity of light in vacuum
v=Velocity of light in medium
Answer:
316.5 cals.
Explanation:
(a) heat needed to change temperature from absolute zero to melting temperature of ice
= mass x specific heat x rise in temperature
= 1 x .5 x 273 = 136.5 cals
b ) heat needed to melt the ice
= mass x latent heat of melting
= 1 x 80 = 80 cals
c ) heat needed to change temperature from melting temperature to boiling temperature.
mass x specific heat x rise in temperature
1 x 1 x 100
= 100 cals
Total heat required
= 136.5 + 80 + 100
= 316.5 cals.