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3 years ago

Please help ty! I added extra points.

Mathematics

1 answer:

shepuryov [24]3 years ago

8 0

<h3>Charlene left city A at 10: 10 AM</h3>

Solution:

Given that,

At 10 a.m., Hillary left City A and drove at an average speed of 120 km/h

Charlene drove at an average speed of 144 km/h and took 50 minutes.

She arrived at City B at the same time as Hillary

Their distance is same

Find the ratio of speed

120 : 144

Reduce to lowest term

Hillary : charlne = 5 : 6

Find the ratio of time:

Since, speed and time are inversely propotional

Hillary : charlne = 6 : 5

Let time taken by Hiliary be 6t

Let time taken by charlene be 5t

Given that, Charlene took 50 minutes

Therefore,

5t = 50

t = 10

Thus,

time taken by Hiliary = 6t = 6(10) = 60

Thus hiliary took 60 minutes to reach

Find the time Charlene left City A

At 10 a.m., Hillary left City A

Difference between both their time = 60 - 50 = 10 minutes

Thus, Hilairy left city A = 10 am + 10 minutes = 10 : 10 a.m

Thus, Hillary left city A at 10: 10 AM

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$\bar x = 82$

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Step-by-step explanation:

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$n = 12$

$87\ 91\ 86\ 82\ 72\ 91\ 60\ 77\ 80\ 79\ 83\ 96$

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To do this, we simply calculate the sample mean

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$\bar x = \frac{87+ 91+ 86+ 82+72+91+60+77+80+79+83+96}{12}$

$\bar x = \frac{984}{12}$

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Solving (b); Point estimate of standard deviation

To do this, we simply calculate the sample standard deviation

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$\sigma = \sqrt\frac{(87-82)^2+ (91-82)^2+ ....+ (79-82)^2+ (83-82)^2+ (96-82)^2}{12-1}$

$\sigma = \sqrt\frac{1022}{11}$

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$\sigma = 9.64$

Note that: The sample mean and the sample standard deviation are the best point estimators for the mean and the standard deviation, respectively.

Hence, the need to solve for sample mean and sample standard deviation

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3 years ago