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Ivahew [28]
3 years ago
11

What point would be reflected across the y-axis to ( − 3, − 9) ?

Mathematics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

(-3, 9)

Step-by-step explanation:

If a point was reflected across the y-axis, then the y-coordinate would change signs (from negative to positive or from positive to negative). Using this, we could find the original point by changing the sign of the y-coordinate. As you may already see, this is really just reflecting the point across the y-axis again. By doing this, you get the point (-3, 9).

The point would be (-3, 9).

I hope this helps. :)

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Given the probability distributions shown to the​ right, complete the following parts.
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a) Expected Value for distribution A, E(X) = 3.020

Expected Value for distribution B, E(X) = 0.980

b) Standard deviation of distribution A = 1.157

Standard deviation of distribution B = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

Step-by-step explanation:

Expected values is given as

E(X) = Σ xᵢpᵢ

where xᵢ = each possible sample space

pᵢ = P(X=xᵢ) = probability of each sample space occurring.

Distributions A and B is given by

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

For distribution A

E(X) = Σ xᵢpᵢ = (0×0.04) + (1×0.09) + (2×0.15) + (3×0.25) + (4×0.47) = 3.02

For distribution B

E(X) = Σ xᵢpᵢ = (0×0.47) + (1×0.25) + (2×0.15) + (3×0.09) + (4×0.04) = 0.98

b) Standard deviation = √(variance)

But Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

For distribution A

Σx²p = (0²×0.04) + (1²×0.09) + (2²×0.15) + (3²×0.25) + (4²×0.47) = 10.46

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Standard deviation = √(1.3396) = 1.157

For distribution B

Σx²p = (0²×0.47) + (1²×0.25) + (2²×0.15) + (3²×0.09) + (4²×0.04) = 2.30

Variance = Var(X) = 2.30 - 0.98² = 1.3396

Standard deviation = √(1.3396) = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

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