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2 years ago

Mei and Anju are sitting next to each other on different horses on a carousel. Mei's horse is 3 meters from the center of the

Mathematics

2 answers:

Law Incorporation [45]2 years ago

6 0

Answer:

The answer below me is correct, He just didnt have answers to go off, of the answer is 2pi because 2 x 3.14 is 6.28

Post answers next time

SOVA2 [1]2 years ago

5 0

Answer:

$2\pi \approx 6.28\ meters$

Step-by-step explanation:

Mei's horse is 3 meters from the center of the carousel. After one rotation, Mei travelled

$l_M=2\pi r=2\pi \cdot 3=6\pi\ meters$

Anju's horse is 2 meters from the center. After one rotation, Mei travelled

$l_A=2\pi r=2\pi \cdot 2=4\pi\ meters$

The difference in their travelled distances is

$l_M-l_A=6\pi -4\pi =2\pi \approx 6.28\ meters$

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An experienced carpenter can frame a house twice as fast as an apprentice. Working together, it takes the carpenters 2 days. How

AleksAgata [21]

Answer:

6 days

Step-by-step explanation:

Let the time taken by Carpenter working alone = $C$ days

Then time taken by apprentice alone = Twice as that of taken by Carpenter = 2 $C$ days

Time taken working together = 2 days

Work done in one day working together = $\frac{1}{2}$

Work done in one day by Carpenter working alone = $\frac{1}{C}$

Work done in one day by apprentice working alone = $\frac{1}{2C}$

Work done in one day by Carpenter working alone + Work done in one day by Carpenter working alone = $\frac{1}{C}$ + $\frac{1}{2C}$ = Work done in one day working together = $\frac{1}{2}$

$\dfrac{1}{C}+\dfrac{1}{2C}=\dfrac{1}{2}\\\Rightarrow \dfrac{2+1}{2C}=\dfrac{1}{2}\\\Rightarrow C = 3\ days$

Time taken by Carpenter alone to complete the work = 3 days

Time taken by Apprentice alone to complete the work = 3 $\times$ 2= 6 days

4 0

2 years ago

The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of c

boyakko [2]

Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063

Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

$p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}$ for x = 0, 1, 2, ...

where $\lambda$ , the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is

$(0.36143)(0.36143) = 0.13063$

4 0

2 years ago

From Environment Canada: "Our experts on weather and climate have combed through 45 years of weather records, and determined the

miskamm [114]

Hat grade are u? I'm just asking cause the formulas are going with grade I could help if u want me too

7 0

2 years ago

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The system of equations is solved using the linear combination method. StartLayout 1st row 1st column one-half x + 4 y = 8 right

Marina CMI [18]

Answer:

it's (A) - There are no solutions to the system because the equations represent parallel lines. Just took the quiz.

Step-by-step explanation:

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2 years ago

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Why is customer awareness a synonym of customer responsibility?

LUCKY_DIMON [66]

Customer awareness is the same as customer responsibility. this is because a customer should always be aware of any deals, defectives, or flaws in the product. being “aware” means “having knowledge of a situation”. being aware of your purchases affects your responsibility and attentiveness. customers should always examine the closely before purchasing.

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2 years ago