The sum of 1 2 and 1 5 is between 1 2 and 1. YES OR NO?

Mathematics

1 answer:

Nastasia [14]3 years ago

6 0

No, the sum of 12 and 15 is 27

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Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distribute

mafiozo [28]

Answer: 1

Step-by-step explanation:

If a random variable x is uniformly distributed in [a,b] the

Mean = $\dfrac{a+b}{2}$

Standard deviation : $\sqrt{\dfrac{(b-a)^2}{12}}$

Let x = Times required for a cable company to fix cable problems

As per given.

x is uniformly distributed between 40 minutes and 65 minutes.

Then , mean = $\dfrac{65+40}{2}=52.5$ minutes

Standard deviation : $\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22$ minutes

Consider , P (mean- 2(Standard deviation) < X < mean+2(Standard deviation) )

= P(52.5-2(7.22)< X < 52.5+2(7.22))

=P(38.06 <X < 66.94 ).

But x lies between 40 minutes and 65 minutes.

Also, [40 minutes, 65 minutes]⊂ [38.06 minutes , 66.94 minutes]

Therefore ,P(38.06 <X < 66.94 ) =1

∴ The probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean is 1.

8 0

3 years ago

Josh is unpacking boxes at her job at target she unpacks 17 boxes that each have 25 regular DVDs in them she also on tax 24 boxe

Musya8 [376]

Answer:

1+1=4

Step-by-step explanation:

6 0

3 years ago

In the expansion (ax+by)^7, the coefficients of the first two terms are 128 and -224, respectively. Find the values of a and b

madam [21]

Answer:

a = 2, b = 3.5

Step-by-step explanation:

Expanding $(ax+by)^7$ using Binomial expansion, we have that:

$(ax+by)^7$ =

$(ax)^7(by)^0 + (ax)^6(by)^1 + (ax)^5(by)^2 + (ax)^4(by)^3 + (ax)^3(by)^4 + (ax)^2(by)^5 + (ax)^1(by)^6 + (ax)^0(by)^7$

$= (a)^7(x)^7+ (a)^6(x)^6(b)(y) + (a)^5(x)^5(b)^2(y)^2 + (a)^4(x)^4(b)^3(y)^3 + (a)^3(x)^3(b)^4(y)^4 + (a)^2(x)^2(b)^5(y)^5 + (a)(x)(b)^6(y)^6 + (b)^7(y)^7\\\\\\= (a)^7(x)^7+ (a)^6(b)(x)^6(y) + (a)^5(b)^2(x)^5(y)^2 + (a)^4(b)^3(x)^4(y)^3 + (a)^3(b)^4(x)^3(y)^4 + (a)^2(b)^5(x)^2(y)^5 + (a)(b)^6(x)(y)^6 + (b)^7(y)^7$