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3 years ago

A cruise ship is traveling at a speed of

Mathematics

1 answer:

Art [367]3 years ago

6 0

Answer:

1 mile = 1760 yards

20 * 1.151 = 23.02

23.02 * 1760 = 40,515.2

40,515.2 / 60 = 675.243~

675.243~ / 60 = 11.2542~

They are traveling 11.25 yards per second

I took the number of knots multiplied by the approximate miles per hour. Then multiplied that by the number of yards in a mile, and then divided that by the number of minutes in an hour, and divided that by the number of seconds in a minute. That is how I got the answer!

when rounded to tens, the answer is 11

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A one compartment vertical file is to be constructed by bending the long side of a 6 in. by 10 in. sheet of plastic along two li

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3 years ago

Please help me with this . I can’t fail geometry I need your help please<br> Please

krok68 [10]

Answer:

Step-by-step explanation:

m<ACB is half of the angle AOB so u divide 62/2 and the answer is31

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3 years ago

Let X denote the temperature (degree C) and let Y denote thetime in minutes that it takes for the diesel engine on anautomobile

BlackZzzverrR [31]

Answer:

Step-by-step explanation:

Given $f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y$

we know that $\int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1$

therefore $\int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1$

on integrating we get

c=(1/6640)

$P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy$

on doing the integration we get

=0.37349

marginal density of X is

$f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy$

on doing integration we get

f(x)=(4x+3)/3320 ; 0<x<40

marginal density of Y is

$f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx$

on doing integration we get

$f(y)=\frac{(y+40.5)}{83}$

$P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy$

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solve the above integration we get the answer

Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.

we know f(x,y)

In the (c) bit we got f(x) and f(y)

f(x,y)cramster-equation-2006112927536330036287f(x).f(y)

therefore X and Y are not independent

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3 years ago

A random sample of 15 employees was selected. The average age in the sample was 31 years with a variance of 49 years. Assuming a

vova2212 [387]

Answer:

Assuming ages are normally distributed, the 98% confidence interval for the population average age is [26.3, 35.7].

Step-by-step explanation:

We have to construct a 98% confidence interval for the mean.

The information we have is:

- Sample mean: 31

- Variance: 49

- Sample size: 15

- The age is normally distributed.

We know that the degrees of freedom are

$k=n-1=15-1=14$

Then, the t-value for a 98% CI is t=2.625 (according to the t-table).

The standard deviation can be estimated from the variance as:

$s=\sqrt {s^2}=\sqrt{49}=7$

The margin of error is:

$E=t_{14}*s/\sqrt{n}\\\\E=2.625*7/\sqrt{15}=18.375/3.873=4.7$

Then, the CI can be constructed as:

$M-t*s/\sqrt{n}\leq\mu\leq M+t*s/\sqrt{n}\\\\31-4.7\leq \mu \leq 31+4.7\\\\26.3\leq \mu \leq 35.7$

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3 years ago

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victus00 [196]

Answer:

3y(x+2z)

Step-by-step explanation:

hope this helps

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3 years ago