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2 years ago

9. What is the likelihood that a randomly selected couple's relationship survived between 90 and 150 days

Mathematics

1 answer:

andrew-mc [135]2 years ago

7 0

Answer:

0.5818

Step-by-step explanation:

Suppose the average teenage romantic relationship is normally distributed with a mean number of 100 days with a standard deviation of 30 days.

Answer:

The z score is used to determine how many standard deviations that the raw score is above or below the mean. If the z score is positive then the raw score is above the mean and if it is negative then it is below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}\\ \\where\ \mu=mean,\sigma=standard\ deviation, x=raw\ score$

Given that:

$\mu=100\ days,\sigma=30\ days\\\\For\ x=90\\\\z=\frac{90-100}{30} =-0.33\\\\For\ x=150\\\\z=\frac{150-100}{30} =1.67$

Therefore, from the normal distribution table, P(90 < x < 150) = P(-0.33 < z < 1.67) = P(z < 1.67) - P(z < -0.33) = 0.9525 - 0.3707 = 0.5818

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serg [7]

Answer:

If Miles sells 6 tickets the money he will collect be $36.

Step-by-step explanation:

Given:

Miles sells 3 raffle tickets for $18 and 4 tickets for $24.

The pattern for the price of the tickets is the same.

Now, to find the money miles will collect if he sells 6 tickets.

Let the money Miles will collect be $x.$

The ticket if sells will be 6.

As, Miles sells 3 tickets for $18.

And the pattern for the price is same.

So, 3 is equivalent to 18.

And, 6 is equivalent to $x.$

Now, to get the money by using cross multiplication method:

$\frac{3}{18} =\frac{6}{x}$

By cross multiplying we get"

$3x=108$

Dividing both sides by 3 we get:

$x=36.$

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Therefore, if Miles sells 6 tickets the money he will collect is $36.

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3 years ago

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Mnenie [13.5K]

Answer:

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

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Step-by-step explanation:

For this case we have the following differential equationÑ

$\frac{du}{dt}= -k (u-T)$

We can reorder the expression like this:

$\frac{du}{u-T} = -k dt$

We can use the substitution $w = u-T$ and $dw =du$ so then we have:

$\frac{dw}{w} =-k dt$

IF we integrate both sides we got:

$ln |w| = -kt +C$

If we apply exponential in both sides we got:

$w = e^{-kt} *e^c$

And if we replace w = u-T we got:

$u(t)= T + C_1 e^{-kt}$

We can also express the solution in the following terms:

$u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}$

For this case we know that $k =-0.15 hr$ since w ehave a cooloing, $T_{i}= 70 F, T_{amb}=11F$ , we have this model:

$u(t) = (70-11) e^{-0.15t} +11$

And if we want that the temperature would be 32F we can solve for t like this:

$32 = 59 e^{-0.15 t} +11$

$21=59 e^{-0.15 t}$

$\frac{21}{59} = e^{-0.15 t}$

If we apply natural logs on both sides we got:

$ln (\frac{21}{59}) =-0.15 t$

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

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