Answer:
90°
Step-by-step explanation:
since PQ is the diameter of the circle, m <em>L</em><em> </em>PRQ = ½ × 180° = 90°.
so, as a supplementary to <em>L</em><em> </em>PRQ, the value of angle QRS = 180 - m <em>L</em><em> </em>PRQ = 180-90 =90°
Answer:
The total cost of 1 belt and 1 tie is $13 and $39 respectively.
Step-by-step explanation:
Given that,
The cost of 4 belts and 5 ties is $247.
Each tie costs 3 times as much as a belt.
Let the cost of a belt is x and that of a tie is y.
ATQ,
4x + 5y = 247 ...(1)
y = 3x ....(2)
Put the value of y from equation (2) in equation (1)
4x + 5(3x) = 247
4x + 15x = 247
19x = 247
x = 13
Put the value in equation (2)
y = 3x
= 3(13)
= 39
So, the total cost of 1 belt and 1 tie is $13 and $39 respectively.
Answer:
The critical value that should be used in constructing the interval is T = 5.8408.
The 99% confidence interval for the true mean yield is between 2.943 bushels per acre and 96.268 bushels per acre.
Step-by-step explanation:
We have the standard deviation of the sample, so we use the students t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 4 - 1 = 3
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of
. So we have T = 5.8408. This is the critical value.
The margin of error is:
M = T*s = 5.8408*7.99 = 46.668 bushels per acre
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 49.6 - 46.668 = 2.943 bushels per acre.
The upper end of the interval is the sample mean added to M. So it is 49.6 + 46.668 = 96.268 bushels per acre.
The 99% confidence interval for the true mean yield is between 2.943 bushels per acre and 96.268 bushels per acre.
Answer:
101, 3, 0.025, 0.7416
Step-by-step explanation:
Given that a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.
As per central limit theorem we have
a) Mean of sample mean = 
Std deviation of sample mean = 
Mean = 101
Std dev =3.00
b) 
c) the probability that the sample mean deviates from the population mean μ = 101 by no more than 4.
=
I found out that t = 117 over 5