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3 years ago

What is the difference between infinite solutions and all real numbers?

Mathematics

1 answer:

igomit [66]3 years ago

6 0

Answer:

Real numbers are like pi, 4, -3, 1/3, 1.25, -3468374, ......

Infinite solutions are like when you solve a problem and it doesn't work out.

like 2+2 = 3 + 2

4 = 5

It doesn't make sense at all

Hope it helps :)

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Assume the sample variances to be continuous measurements. Find the probability that a random sample of 25observations, from a n

konstantin123 [22]

Answer:

a) $P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

b) $P(13.848$ Step-by-step explanation:

Previous concepts

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

For this case we assume that the sample variance is given by $S^2$ and we select a random sample of size n from a normal population with a population variance $\sigma^2$ . And we define the following statistic:

$T = \frac{(n-1) S^2}{\sigma^2}$

And the distribution for this statistic is $T \sim \chi^2_{n-1}$

For this case we know that n =25 and $\sigma^2 = 6$ so then our statistic would be given by:

$\chi^2 = \frac{(n-1)S^2}{\sigma^2}=\frac{24 S^2}{6}= 4S^2$

With 25-1 =24 degrees of freedom.

Solution to the problem

Part a

For this case we want this probability:

$P(S^2 > 9.1)$

And we can multiply the inequality by 4 on both sides and we got:

$P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

And we can use the following excel code to find it: "=1-CHISQ.DIST(36.4,24,TRUE)"

Part b

For this case we want this probability:

$P(3.462 < S^2$

If we multiply the inequality by 4 on all the terms we got:

$P(3.462*4 < 4S^2 < 4*10.745)= P(13.848< \chi^2$ And we can find this probability like this: $P(13.848$ And we use the following code to find the answer in excel: "=CHISQ.DIST(42.98,24,TRUE)-CHISQ.DIST(13.848,24,TRUE)"

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3 years ago

Guys please help i’m blanking!!!<br> Select all the functions whose graphs include the point (16,4)

Gnom [1K]

D and E —> substitute the x or y of the point given into the functions to test if it is correct

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3 years ago

I feel like I am asking so my questions but I dont really care its a Sunday xD

dimulka [17.4K]

You would add like terms so it would be 1x then you add 1 to 7 so you would get 8.. the you divide both sides by 1x and you'd get 8=x

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3 years ago

Read 2 more answers

Hi again :') i got a new question

const2013 [10]

The answer to your question is 9 x 9 = 81.

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3 years ago

Read 2 more answers

Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without b

Korolek [52]

Hello,

Very nice as problem.

2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies

since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0

4 0

3 years ago