<span>If all your entries are integers, the determinant is also an integer.
</span>For this problem, 1 0
0 13
For this question, you should just fix 3 values and make the 4th value an unknown and solve it.
I hope my answer helped you somehow.
Answer: 0.2551
Step-by-step explanation:
Given : The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the actual temperature of the medium, and standard deviation σ.
Significance level : 
The critical z-value for 95% confidence :
(1)
Since ,
(where x be any random variable that represents the temperature reading from a thermocouple.)
Then, from (1)
(2)
Also, all readings are within 0.5° of μ,
i.e. 
i.e.
[From (2)]
i.e.
i.e.
The required standard deviation : 
To solve this problem you must follow the proccedure shown below:
1. Amount of lemonade in pints:
1 <span>quart of water=2 pints
1 pint of lemon
(9 ounces of honey)(0.0625 pints/1 ounce)=0.56 pints
Total in pints=2 pints+1 pint+0.56 pints
Total in pints=3.56 pints
2. </span>Amount of lemonade in<span> cups:
Total in cups=(3.56 pints)(2 cups/1 pint)
Total in cups=7.12 cups
3. </span>Amount of lemonade in<span> ounces:
Total in ounces=(7.12 cups)(8 ounces/1 cup)
Total in ounces=56.96 ounces</span>
5/3 = 1.66667
1.666666^3=4.6296 (Volume is measured in cubic units)
54x4.6396=250
250 x pi = 250Pi.
Answer:
The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.
Step-by-step explanation:
What is Confidence Interval?
The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.
The confidence interval is given by

Where
is the mean weight
is the standard deviation
is the critical value from t-table and n is the sample size.
The term
is known as margin of error.
As the sample size is decreased the corresponding margin of error increases which results in wider confidence interval which means smaller precision.
The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.
We can say with 95% confidence that the true mean weight of the rock is within the interval of (25.2, 29.1).