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ANEK [815]
3 years ago
12

Connor rolled a die 6 times. A 2 occurred three times. Why isn't the theoretical probability 3/6 instead of 1/6?​

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
6 0

2 occurred 3times

\\ \sf\longmapsto S=6\times 3=18

\\ \sf\longmapsto |S|=18

Now

\\ \sf\longmapsto E=No\:of\:times\: 2\:occured

\\ \sf\longmapsto |E|=3

we know

\boxed{\sf P(E)=\dfrac{|E|}{|S|}}

\\ \sf\longmapsto P(E)=\dfrac{3}{18}

\\ \sf\longmapsto P(E)=\dfrac{1}{6}

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(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

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4 years ago
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Step-by-step explanation:

64+36=100

8 squared plus 6 squared equals 10 squared

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Twenty-five students from Harry High School were accepted at Magic University. Of those students, 10 were offered athletic schol
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Answer:

Step-by-step explanation:

Part A

For Athletic scholarship,

Mean = (16 + 24 + 20 + 25 + 24 + 23 + 21 + 22 + 20 + 20)/10 = 21.5

Standard deviation = √(summation(x - mean)²/n

n = 10

Summation(x - mean)² = (16 - 21.5)^2 + (24 - 21.5)^2 + (20 - 21.5)^2 + (25 - 21.5)^2 + (24 - 21.5)^2 + (23 - 21.5)^2 + (21 - 21.5)^2 + (22 - 21.5)^2 + (20 - 21.5)^2 + (20 - 21.5)^2 = 64.5

Standard deviation = √64.5/10 = 2.54

For non athletic scholarship,

Mean = (23 + 25 + 26 + 30 + 32 + 26 + 28 + 29 + 26 + 27 + 29 + 27 + 22 + 24 + 25)/15 = 26.6

n = 15

Summation(x - mean)² = (23 - 26.6)^2 + (25 - 26.6)^2 + (26 - 26.6)^2 + (30 - 26.6)^2 + (32 - 26.6)^2 + (26 - 26.6)^2 + (28 - 26.6)^2 + (29 - 26.6)^2 + (26 - 26.6)^2 + (27 - 26.6)^2 + (29 - 26.6)^2 + (27 - 26.6)^2 + (22 - 26.6)^2 + (24 - 26.6)^2 + (25 - 26.6)^2 = 101.6

Standard deviation = √101.6/15 = 2.6

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let 1 be the subscript for scores of athletes and 2 be the subscript for scores of non athletes.

Therefore, the population means would be μ1 and μ2

The random variable is x1 - x2 = difference in the sample mean scores of athletes and non athletes.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 21.5

x2 = 26.6

s1 = 2.54

s2 = 2.6

n1 = 10

n2 = 15

t = (21.5 - 26.6)/√(2.54²/10 + 2.6²/15)

t = - 4.65

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [2.54²/10 + 2.6²/15]²/[(1/10 - 1)(2.54²/10)² + (1/15 - 1)(2.6²/15)²] = 1.2008/0.1039

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p value = 0.00056

Since alpha, 0.1 > than the p value, 0.00056, then we would reject the null hypothesis.

Therefore, these data provide convincing evidence of a difference in ACT scores between athletes and nonathletes.

Part B

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 90% confidence level, the z score from the normal distribution table is 1.645

x1 - x2 = 21.5 - 26.6 = - 5.1

√(s1²/n1 + s2²/n2) = √(2.54²/10 + 2.6²/15) = 1.05

The confidence interval is - 5.1 ± 1.05

This analysis provides evidence that the mean scores for non athletes is higher than the mean scores for athletes, and that the difference between means in the population is likely to be between 4.05 and 6.15

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