Please help I don't understand

Mathematics

1 answer:

DIA [1.3K]2 years ago

5 0

Answer:

14/18

Step-by-step explanation:

7/9 x 2= 14/18

7/9=14/18

but not sure what proportion means but just look up definition and yeah

I hope this was helpful!

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1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie

Ad libitum [116K]

Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

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The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

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Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

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At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

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At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21

5 0

3 years ago

Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of

Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

$\triangle ABC \sim \triangle NMP$