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zysi [14]
3 years ago
11

Suppose the discrete random variable X has the probability distribution below:

Mathematics
1 answer:
Talja [164]3 years ago
4 0

(a) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.11 + 0.52 + 0.19 = 0.82

(b) P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.52 + 0.19 + 0.12 + 0.06 = 0.89

(c) µ = 0×0.11 + 1×0.52 + 2×0.19 + 3×0.12 + 4×0.06 = 1.5

(d) σ² = (0²×0.11 + 1²×0.52 + 2²×0.19 + 3²×0.12 + 4²×0.06) - µ² = 1.07

σ = √(σ²) ≈ 1.03

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IM GIVING BRAINLIEST!!!PLEASE HELP!!!I JUST KNOW ITS NOT A!!!
zmey [24]

Answer:

C. In y^2-y=6, 6 should have been subtracted on both sides first.

Step-by-step explanation:

It is a quadratic equation so you subtract the 6 and create

y^2-y-6=0

Then you factor it to:

(y-3)(y+2)=0

The solutions then are:

y-3=0, y=3 and y+2=0, y=-2

{-2,3}

4 0
3 years ago
Someone pls help i need help with my math question
11111nata11111 [884]

Answer:

21 units

Step-by-step explanation:

6 0
2 years ago
Can someone help me figure out 43-45!please
Irina18 [472]
43) ang 1 = ang 2
5x = 9(x-4)
5x = 9x - 36
4x = 36
x = 9. so ang 1 + ang 2 = 90°

44) ang 1 +ang 2 + ang 3 = 180
60 + 7x+1 + 9x+7 = 180
16x = 112
x = 7
so angle 2 is 7*7+1= 50°
angle 3 is 9*7+7= 70°

45) let ang 2=x, then ang 1 is 3x
then 3x+x=96
4x = 96
x=24
angle 2 is 24° and angle 1 is 72°

8 0
3 years ago
7th grade sac practice problems look at the triangular prism below. each triangular face of the prism has a base of 3 centimeter
xxMikexx [17]

Answer:

72cm³

Step-by-step explanation:

Base of the triangular face =3 cm.

Height of the triangular face = 4 cm.

Length of the prism = 12 cm.

Volume of a Prism= Base Area X. length

The base of a triangle prism is a triangle, therefore:

Volume of a Triangular Prism= ½bh X Length

=½ X 3 X 4 X 12

=½ X 144

=72cm³

Volume of the triangular Prism= 72cm³

4 0
3 years ago
Suppose parts (a) through (d) below provide results for a study on the role of calcium in reducing the symptoms of PMS. For each
frutty [35]

Answer:

Step-by-step explanation:

Hello!

To test if calcium reduces the symptoms of PMS two independent groups of individuals are compared, the first group, control, is treated with the placebo, and the second group is treated with calcium.

The parameter to be estimated is the difference between the mean symptom scores of the placebo and calcium groups, symbolically: μ₁ - μ₂

There is no information about the distribution of both populations X₁~? and X₂~? but since both samples are big enough, n₁= 228 and n₂= 212, you can apply the central limit theorem and approximate the sampling distribution to normal X[bar]₁≈N(μ₁;δ₁²/n) and X[bar]₂≈N(μ₂;δ₂²/n)

The formula for the CI is:

[(X[bar]₁-X[bar]₂) ± Z_{1-\alpha /2} * \sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} }]

95% confidence level Z_{1-\alpha /2}= Z_{0.975}= 1.96

(a) mood swings: placebo = 0.70 ± 0.78; calcium = 0.50 ± 0.53

X₁: Mood swings score of a participant of the placebo group.

X₂: Mood swings score of a participant of the calcium group.

[(0.70-0.50) ± 1.96 * \sqrt{\frac{0.78^2}{228} +\frac{0.53^2}{212} }]

[0.076; 0.324]

(b) crying spells: placebo = 0.39 + 0.57; calcium = 0.21 + 0.40

X₁: Crying spells score of a participant of the placebo group.

X₂: Crying spells score of a participant of the calcium group.

[(0.39-0.21) ± 1.96 * \sqrt{\frac{0.57^2}{228} +\frac{0.40^2}{212} }]

[0.088; 0.272]

(c) aches and pains: placebo = 0.45 + 0.60; calcium = 0.37 + 0.45

X₁: Aches and pains score of a participant of the placebo group.

X₂: Aches and pains score of a participant of the calcium group.

[(0.45-0.37) ± 1.96 * \sqrt{\frac{0.60^2}{228} +\frac{0.45^2}{212} }]

[-0.019; 0.179]

(d) craving sweets or salts: placebo = 0.60 + 0.75; calcium = 0.44 + 0.61

X₁: Craving for sweets or salts score of a participant of the placebo group.

X₂: Craving for sweets or salts score of a participant of the calcium group.

[(0.60-0.44) ± 1.96 * \sqrt{\frac{0.75^2}{228} +\frac{0.61^2}{212} }]

[0.032; 0.287]

I hope this helps!

6 0
3 years ago
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