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Mariana [72]
3 years ago
11

Suppose that you drop a ball from a window 50 metres above the ground. The ball bounces to 50% of its previous height with each

bounce. If the ball continues to bounce in this manner how far will it have traveled, up and down, from the time it was dropped from the window until the 25th bounce?
200 m
100 m
75 m
150 m
Mathematics
1 answer:
stich3 [128]3 years ago
4 0

We have been given that  you drop a ball from a window 50 metres above the ground. The ball bounces to 50% of its previous height with each bounce. We are asked to find the total distance traveled by up and down from the time it was dropped from the window until the 25th bounce.            

We will use sum of geometric sequence formula to solve our given problem.

S_n=\frac{a\cdot(1-r^n)}{1-r}, where,

a = First term of sequence,

r = Common ratio,

n = Number of terms.

For our given problem a=50, r=50\%=\frac{50}{100}=0.5 and n=25.

S_{25}=\frac{50\cdot(1-(0.5)^{25})}{1-0.5}

S_{25}=\frac{50\cdot(1-0.0000000298023224)}{0.5}

S_{25}=\frac{50\cdot(0.9999999701976776)}{0.5}

S_{25}=100\cdot(0.9999999701976776)

S_{25}=99.99999701976776\approx 100

Therefore, the ball will travel 100 meters and option B is the correct choice.

 

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zhenek [66]

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96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is  [335.89 , 356.10].

Step-by-step explanation:

We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

           P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = average rent paid by 314 BYU students = $346

            s = sample standard deviation = $86

            n = sample of students = 314

            \mu = population mean monthly rent of all unmarried BYU students

So, 96% confidence interval for the population mean monthly rent, \mu is ;

P(-2.082 < t_3_1_3 < 2.082) = 0.96

P(-2.082 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.082) = 0.96

P( -2.082 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.082 \times {\frac{s}{\sqrt{n} } } ) = 0.96

P( \bar X -2.082 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ) = 0.96

96% confidence interval for \mu = [ \bar X -2.082 \times {\frac{s}{\sqrt{n} } } , \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ]

                                                = [ 346 -2.082 \times {\frac{86}{\sqrt{314} } } , 346 +2.082 \times {\frac{86}{\sqrt{314} } } ]

                                                = [335.89 , 356.10]

Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

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