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2 years ago

6

Loren drove 200 miles at a certain rate, and his wife,Lois, drove 100 miles at a rate 10 mph slower. If Loren had driven for the

entire trip, they would have arrived 30 minuets sooner. What was Loren’s rate?

1 answer:

Elanso [62]2 years ago

8 0

20 MPH I think.
Hope this helps

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Recent census data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. A random sample of 125

kicyunya [14]

Answer:

The probability is $P(14 < X < 20 ) = 0.5354$

Step-by-step explanation:

From the question we are told that

The proportion that live with their parents is $\r p = 0.142$

The sample size is n = 125

Given that there are two possible outcomes and that this outcomes are independent of each other then we can say the Recent census data follows a Binomial distribution

i.e

$X \ \~ \ B( \mu , \sigma )$

Now the mean is evaluated as

$\mu = n * \r p$

$\mu = 125 * 0.142$

$\mu = 17.75$

Generally the proportion that are not staying with parents is

$\r q = 1 - \r p$

= > $\r q = 0.858$

The standard deviation is mathematically evaluated as

$\sigma = \sqrt{n * \r p * \r q }$

$\sigma = \sqrt{ 125 * 0.142 * 0.858 }$

$\sigma = 3.90$

Given the n is large then we can use normal approximation to evaluate the probability as follows

$P(14 < X < 20 ) = P( \frac{ 14 - 17.75}{3.90}$

Now applying continuity correction

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < \frac{ X - \mu }{\sigma } < \frac{ 19.5 - 17.75}{3.90} )$

Generally

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

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < Z< \frac{ 19.5 - 17.75}{3.90} )$

$P(14 < X < 20 ) = P( -1.0897 < Z< 0.449 } )$

$P(14 < X < 20 ) = P( Z< 0.449 ) - P(Z < -1.0897)$

So for the z - table

$P( Z< 0.449 ) = 0.67328$

$P(Z < -1.0897) = 0.13792$

$P(14 < X < 20 ) = 0.67328 - 0.13792$

$P(14 < X < 20 ) = 0.5354$

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Dahasolnce [82]

7x^2 - x = 6
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x - 1 = 0; x = 1

answer
x = -6/7; x = 1

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X = 9/7. 9/x + 9/x = 14 18/x = 14 18 = 14x 18/14 = x 9/7 = x

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