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Answer:
The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.
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 = 12 - 1 = 11
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.201
The margin of error is:
In which s is the standard deviation of the sample(square root of the variance) and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 19.2 - 1.3 = 17.9 students.
The upper end of the interval is the sample mean added to M. So it is 19.2 + 1.3 = 20.5 students
The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.
Answer:
Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.
To prove: Point A,D,P are collinear.
Proof:
→Case 1. When vertices A and P are opposite side of Base BC.
In Δ ABD and Δ ACD
AB= AC [Given]
AD is common.
BD=DC [median of a triangle divides the side in two equal parts]
Δ ABD ≅Δ ACD [SSS]
∠1=∠2 [CPCT].........................(1)
Similarly, Δ PBD ≅ Δ PCD [By SSS]
∠ 3 = ∠4 [CPCT].................(2)
But, ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]
2 ∠2 + 2∠ 4=360° [using (1) and (2)]
∠2 + ∠ 4=180°
But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.
So, point A, D, P lies on a line.
CASE 2.
When ΔABC and ΔPBC lie on same side of Base BC.
In ΔPBD and ΔPCD
PB=PC[given]
PD is common.
BD =DC [Median of a triangle divides the side in two equal parts]
ΔPBD ≅ ΔPCD [SSS]
∠PDB=∠PDC [CPCT]
Similarly, By proving ΔADB≅ΔADC we will get, ∠ADB=∠ADC[CPCT]
As PD and AD are medians to same base BC of ΔPBC and ΔABC.
∴ P,A,D lie on a line i.e they are collinear.
