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3 years ago

5

John took a 5 1/2 mile walk to his friend's house. He left at 11 a.M. And arrived at his friend's house at 12:45 p.M. Therefore

it took him 1 3/4 hours to get to his friends house. If the return trip took 2 1/4 hours, what is the average speed on the return trip?

1 answer:

forsale [732]3 years ago

6 0

Answer:

The average speed on the return trip is 2.44 miles/h.

Step-by-step explanation:

To find the average speed we need to use the following equation:

$\overline{v} = \frac{d}{t}$

Where:

d: is the distance

t: is the time

Since the question is to find the average speed on the return trip, we will take the data of the time and speed of only the return trip.

$\overline{v} = \frac{d}{t} = \frac{5 + 1/2}{2 + 1/4} = 2.44 miles/h$

Therefore, the average speed on the return trip is 2.44 miles/h.

I hope it helps you!

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If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal

lorasvet [3.4K]

Answer:

a

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

b

The probability is $P(Z > 2.5 ) = 0.0062097$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 800$

The variance is $var(x) = 1600 \ kg$

The range consider is $x_1 = 778 \ kg \ x_2 = 834 \ kg$

The value consider in second question is $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{var (x)}$

substituting value

$\sigma = \sqrt{1600}$

$\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as

$P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

Generally $\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)$

So

$P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

$P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table the value for $P(z_1 < 0.85) = 0.80234$

and $P(z_1 < -0.55) = 0.29116$

So

$P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

$P(x_1 < X < x_2 ) = 0.51118$

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

$P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

$P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

$P(X > x ) = P(Z >2.5 )$

From the z-table the value for $P(Z > 2.5 ) = 0.0062097$

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Answer:

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Step-by-step explanation:

Open the file, it is solved for you, your welcome.

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Answer:

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STEP 2

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STEP 3

Add the squared answers in STEP 2 then find the mean

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STEP 4

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