Answer:Transform the equation to isolate x: ax = bx + 1. How is the value of x related to the difference of a and b
Step-by-step explanation:
Answer:
Step-by-step explanation:
number of samples, n = 10
Mean = (48 + 51 + 46 + 52 + 47 + 48 + 47 + 50 + 51 + 59)/10 = 49.9
Standard deviation = √(summation(x - mean)/n
Summation(x - mean) = (48 - 49.9)^2 + (51 - 49.9)^2 + (46 - 49.9)^2+ (52 - 49.9)^2 + (47 - 49.9)^2 + (48 - 49.9)^2 + (47 - 49.9)^2 + (50 - 49.9)^2 + (51 - 49.9)^2 + (59- 49.9)^2 = 128.9
Standard deviation = √128.9/10 = 3.59
Confidence interval is written in the form,
(Sample mean - margin of error, sample mean + margin of error)
The sample mean, x is the point estimate for the population mean.
Margin of error = z × s/√n
Where
s = sample standard deviation
From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score
In order to use the t distribution, we would determine the degree of freedom, df for the sample.
df = n - 1 = 10 - 1 = 9
Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05
α/2 = 0.05/2 = 0.025
the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975
Looking at the t distribution table,
z = 2.262
Margin of error = 2.262 × 3.59/√10
= 2.57
the lower limit of this confidence interval is
49.9 - 2.57 = 47.33
the lower limit of this confidence interval is
49.9 + 2.57 = 52.47
So it is false
Answer:
if im right 1600? i did it in my head so im not 100 percent sure
Step-by-step explanation:
Hey there.
1.) The mode is the number that's most frequent in a given list.
In our list, we have 2, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 9, 10.
Our most frequent number is 6; therefore, our mode is 6.
2.) The median is the number in the middle of a list organized from least to greatest.
2, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 9, 10;
4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 9;
4, 5, 5, 5, 6, 6, 6, 6, 7;
5, 5, 5, 6, 6, 6, 6;
5, 5, 6, 6, 6;
5, 6, 6;
6.
6 is our median.
I hope this helps, despite your answer choices not providing the correct answer.
Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if 
and 
then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.
side note: q = 1-p is the complement of probability p
