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Ber [7]
3 years ago
7

g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20

,50,6) to get one random sample of size 20. Determine the mean and standard deviation from this sample, compare these with the population mean and standard deviation.
Mathematics
1 answer:
Illusion [34]3 years ago
5 0

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

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The graph represented in the figure shows a set of linear equations each of which is represented a straight line.

Step-by-step explanation:

System of Equation can be referred to as an assortment of equations to be dealt with. Common examples include linear equations and non-linear equations such as a parabola, hyperbola etc.

Linear set of equations are the most simple of equation depicting a linear relationship between two variables.

E.g.  Y=4x+3

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Read 2 more answers
[4]
Tanya [424]

Answer:

Angle ACD = 38°

Step-by-step explanation:

The full, correct question is presented the attached image to this solution.

Given

Point O is the centre if the circle

Points A, B, C and D are points on the circle

Angle AOB = 140°

Angle OAC = 14°

Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)

140° = 2 × (Angle ACB)

Angle ACB = (140°/2) = 70°

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]

But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°

140° + a + a = 180°

2a = 40°

a = (40/2) = 20°

Angle OAB = Angle ABO = 20°

Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°

Triangle ADC is an iscosceles triangle, hence,

Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]

But

(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]

Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)

Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)

(Angle DCB) + (Angle DAB) = 180°

(x + 70°) + (x + 34°) = 180°

2x + 104° = 180°

2x = 180° - 104° = 76°

x = (76°/2) = 38°

Angle DAC = Angle ACD = 38°

Angle ACD = 38°

Hope this Helps!!!

7 0
2 years ago
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