Plz help me with this Answer options: Quadratic, exponential, cubic and linear
2 answers:
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Answer:
The 95% confidence interval for the mean birth weight of all non-premature babies is between 3350.4 grams and 3662.4 grams. This means that we are 95% sure that true mean birth weight of all non-premature babies is in this interval.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used 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 = 41 - 1 = 40
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 40 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0211
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 3506.4 - 156 = 3350.4 grams
The upper end of the interval is the sample mean added to M. So it is 3506.4 + 156 = 3662.4 grams
The 95% confidence interval for the mean birth weight of all non-premature babies is between 3350.4 grams and 3662.4 grams. This means that we are 95% sure that true mean birth weight of all non-premature babies is in this interval.
Answer:
Do not qualify
Step-by-step explanation:
Yearly income = $58000
Monthly income = = $4833.33
Median annual income = $47778
Monthly expenses = $4600
Income < Median Income ? 58000 < 47778 -> No
x = (Monthly income - expenses) * 60
= (4833.33 - 4600) * 60
= 233.33 * 60
= 13999.8
x < 10000 ? 13999.8 < 10000 -> No
So, Do not qualify