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yawa3891 [41]
3 years ago
9

Show equation has a slope of -3 and a y-intercept of 8

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
8 0

Answer:

y=-3x+8

Step-by-step explanation:

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A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has
bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113  

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711  

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] =  1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0
3 years ago
(CO 4) In a sample of 8 high school students, they spent an average of 24.8 hours each week doing sports with a sample standard
AveGali [126]

Answer:  

(22.12, 27.48)

Step-by-step explanation:

Given : Significance level : \alpha: 1-0.95=0.05

Sample size : n= 8 , which is a small sample (n<30), so we use t-test.

Critical values using t-distribution: t_{n-1,\alpha/2}=t_{7,0.025}=2.365

Sample mean : \overline{x}=24.8\text{ hours}

Standard deviation : \sigma=3.2\text{ hours}

The confidence interval for population means is given by :-

\overline{x}\pm t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}

i.e. 24.8\pm(2.365)\dfrac{3.2}{\sqrt{8}}

24.8\pm2.67569206001\\\\\approx24.8\pm2.68\\\\=(24.8-2.68, 24.8+2.68)=(22.12, 27.48)

Hence, the 95% confidence interval, assuming the times are normally distributed.=  (22.12, 27.48)

6 0
3 years ago
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