Answer:
Third quartile (Q₃) = 46.75 minutes.
Therefore, Option (c) is the correct answer.
Step-by-step explanation:
Given: Mean (μ) = 40 minutes and S.D (σ) = 10 minutes
To find : Third quartile (Q₃) = ?
Sol: As the third quartile of normal distribution covers the 75% of the total area of the curve and first quartile covers the 25% of the total area of the curve. Then with the help of z score table, the value represented the third quartile of the normal distribution is:
Q₃ = μ + 0.675 σ
Now by substitution the value of mean and standard deviation,
Q₃ = 40 + 0.675 × (10)
Q₃ = 40 + 6.75
Q₃ = 46.75
Therefore, the third quartile (Q₃) = 46.75. So, option (c) is the correct answer.
Answer:
15u + 141 = 420
Step-by-step explanation:
So basically we are modelling how much the coach is spending. Each player is given a uniform and a basketball
We can use 'u' as the cost of a single uniform
We can use 'b' as the cost of a single basketball
Since there is 15 players we must give a ball and uniform to each so
15u + 15b = 420
The problem gives us the cost of each basketball: 9.40
so now 15u + 15(9.4) = 420
15u + 141 = 420
-2n + 6 is the answer to this question
Answer:
option-C
Step-by-step explanation:
We are given
At 9:00 a.m., a wind speed of 20 miles per hour was recorded
So, initial wind speed =20 miles per hour
Each measurement showed an increase in wind speed of 3 miles per hour
The strongest wind was recorded at 4:00 p.m
so,
top wind speed = initial wind speed + ( total number of hours between 9:00 am and 4:00 pm)*(change in wind speed)
top wind speed = 20 mph+(16-9)*3
miles per hour
Since, change in wind speed and initial wind speed are constant
but the number of hours it took for the wind speed to reach its minimum for the day can be changed
so, most important variables is
the number of hours it took for the wind speed to reach its minimum for the day can be changed
Answer:
Hope this helps
Step-by-step explanation:
