What is the answer of 8-15=
2 answers:
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Answer: Lattice parameter, a = (4R)/(√3)
Step-by-step explanation:
The typical arrangement of atoms in a unit cell of BCC is shown in the first attachment.
The second attachment shows how to obtain the value of the diagonal of the base of the unit cell.
If the diagonal of the base of the unit cell = x
(a^2) + (a^2) = (x^2)
x = a(√2)
Then, diagonal across the unit cell (a cube) makes a right angled triangle with one side of the unit cell & the diagonal on the base of the unit cell.
Let the diagonal across the cube be y
Pythagoras theorem,
(a^2) + ((a(√2))^2) = (y^2)
(a^2) + 2(a^2) = (y^2) = 3(a^2)
y = a√3
But the diagonal through the cube = 4R (evident from the image in the first attachment)
y = 4R = a√3
a = (4R)/(√3)
QED!!!
Answer:
The number of bacteria after will be
Step-by-step explanation:
Given the initially 30 bacteria present in the culture.
Also, the number of bacteria got doubled every hour.
So, using the equation
Where is number of bacteria after
hours.
is bacteria present initially.
is the common ration, in our problem it is given that bacteria doubles every hour. So,
And is the number of hours. In our problem we need amount of bacteria at the end of
hours. So,
Plugging values in the formula we get,
So, number of bacteria after will be
Answer:
The value of the test statistic is z = 1.39.
The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.
Step-by-step explanation:
Using traditional methods it takes 92 hours to receive an advanced flying license.
This means that at the null hypothesis, it is tested if the mean is of 92, that is:
Test if there is evidence that the technique lengthens the training time
At the alternative hypothesis, it is tested if the mean is more than 92, that is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
92 is tested at the null hypothesis:
This means that
A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.
This means that
Value of the test statistic:
The value of the test statistic is z = 1.39.
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.
Looking at the z-table, z = 1.39 has a p-value of 0.9177.
1 - 0.9177 = 0.0823.
The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.