Answer: Choice A.) 88.2 < μ < 93.0
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Explanation:
We have this given info:
- n = 92 = sample size
- xbar = 90.6 = sample mean
- sigma = 8.9 = population standard deviation
- C = 99% = confidence level
Because n > 30 and because we know sigma, this allows us to use the Z distribution (aka standard normal distribution).
At 99% confidence, the z critical value is roughly z = 2.576; use a reference sheet, table, or calculator to determine this.
The lower bound of the confidence interval (L) is roughly
L = xbar - z*sigma/sqrt(n)
L = 90.6 - 2.576*8.9/sqrt(92)
L = 88.209757568781
L = 88.2
The upper bound (U) of this confidence interval is roughly
U = xbar + z*sigma/sqrt(n)
U = 90.6 + 2.576*8.9/sqrt(92)
U = 92.990242431219
U = 93.0
Therefore, the confidence interval in the format (L, U) is approximately (88.2, 93.0)
When converted to L < μ < U format, then we get approximately 88.2 < μ < 93.0 which shows that the final answer is choice A.
We're 99% confident that the population mean mu is somewhere between 88.2 and 93.0
Terminating because the decimal ends rather than going on forever which would be repeating
B. 16.30 = 16.3
if there is a 0 in a decimal and it is the farthest right place value it is the same as if the 0 wasn't there
Answer:
k=6
Step-by-step explanation:
Line partition formula
1/b(x2-x1)+x1, 1/b(y2-y1)+y1
Where b is the number partitions.
We know the x values so Subsitue 17 for x2 and 2 for x1. and we know this value must equal 7.
1/b(17-2)+2=7
1/b(15)=5
1/b=1/3
b=3
so the partition is 1/3
So let find the y coordinate
So our y coordinate is 6.
If the point U is between points T and V, then the numerical length of TV is 29 units
<h3>How to determine the numerical length of segment TV?</h3>
From the question, we have the following lengths that can be used in our computation:
- Length TU = 18 units
- Length UV = 11 units
The above parameters and representations implies that the point U is between endpoints T and V
This also means that the length TV is longer than the other lengths TU and TV
So, we have the following length equation
TV = TU + UV
Substitute the known values in the above equation
So, we have the following equation
TV = 18 + 11
Evaluate the sum of the like terms in the above equation
So, we have the following equation
TV = 29
Hence, the numerical length of segment TV is 29 units
Read more about lengths at
brainly.com/question/19131183
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<u>Possible question</u>
If tu = 18 and uv = 11 what is tv, if point u is between points t and v
