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iogann1982 [59]

3 years ago

14

Two trains leave the station at the same time, one heading west and the other east. The westbound train travels at 95 miles per

hour. The eastbound train travels at 105 miles per hour. How long will it take for the two trains to be 440 miles apart?
Do not do any rounding.

[] hours

1 answer:

Ludmilka [50]3 years ago

8 0

Answer:

2.2 hr

Step-by-step explanation:

Let t = time

95t = distance of westbound train

105t = distance of eastbound train

95t + 105t = 440

200t = 440

t = 2.2 hr.

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Answer:

a

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

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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

$\sigma = \sqrt{1600}$

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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

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

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

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

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