Two bowlers bowl the following number of strikes in 9 games

1 answer:

8 0

First, let's calculate the mean and the mean absolute deviation of the first bowler.

FIRST BOWLER: 8,5,5,6,8,7,4,7,6
Mean = (Sum of all data)/(Number of data points) = (8+5+5+6+8+7+4+7+6)/9
Mean = 6.222
Mean absolute deviation or MAD = [∑(|Data Point - Mean|]/Number of Data Points
MAD = [|8 - 6.222| + |5 - 6.222| + |5 - 6.222| + |6 - 6.222| + |8 - 6.222| + |7 - 6.222| + |4 - 6.222| + |7 - 6.222| + |6 - 6.222|]/9
MAD = 1.136

SECOND BOWLER: 10,6,8,8,5,5,6,8,9
Mean = (Sum of all data)/(Number of data points) = (10+6+8+8+5+5+6+8+9)/9
Mean = 7.222
Mean absolute deviation or MAD = [∑(|Data Point - Mean|]/Number of Data Points
MAD = [|10 - 7.222| + |6 - 7.222| + |8 - 7.222| + |8 - 7.222| + |5 - 7.222| + |5 - 7.222| + |6 - 7.222| + |8 - 7.222| + |9 - 7.222|]/9
MAD = 1.531

The mean absolute deviation represents the average distance of each data to the mean. Thus, the lesser the value of the MAD is, the more consistent is the data to the mean. Between the two, the first bowler is more consistent.

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The heat developed in an electric wire varies jointly as the wires resistance, the time the current flows, & the square of t

never [62]

Answer:

4 ohms

Step-by-step explanation:

The question is "In two minutes a current of 5 amps develops 1,200 heat units in a wire of 8 ohms resistance. What resistance does a similar wire have, which develops 6,000 heat units with a current of 10 amps in 5 minutes?"

Current, I₁ = 5 A

Heat, Q₁ = 1200 units

Resistance, R₁ = 8 ohms

Time, t₁ = 2 min = 120 s

New heat, Q₂ = 6000 units

Current, I₂ = 10 A

New time, t₂ = 5 min = 300 s

We need to find the new resistance.

Heat developed is given by :

$Q=I^2Rt$

$\dfrac{Q_1}{I_1^2R_1t_1}=\dfrac{Q_2}{I_2^2R_2t_2}\\\\R_2=\dfrac{Q_2\times I_1^2R_1t_1}{Q_1I_2^2t_2}\\\\R_2=\dfrac{6000\times 5^2\times 8\times 120}{1200\times 10^2\times 300}\\\\=4\ \Omega$