We need to 'standardise' the value of X = 14.4 by first calculating the z-score then look up on the z-table for the p-value (which is the probability)
The formula for z-score:
z = (X-μ) ÷ σ
Where
X = 14.4
μ = the average mean = 18
σ = the standard deviation = 1.2
Substitute these value into the formula
z-score = (14.4 - 18) ÷ 1..2 = -3
We are looking to find P(Z < -3)
The table attached conveniently gives us the value of P(Z < -3) but if you only have the table that read p-value to the left of positive z, then the trick is to do:
1 - P(Z<3)
From the table
P(Z < -3) = 0.0013
The probability of the runners have times less than 14.4 secs is 0.0013 = 0.13%
Answer:
option 3 is the answer of this
Answer:
Age of Alice = 25, Age of Jane = 20.
Step-by-step explanation:
Let age of Alice = x
Then age of Jane is 80% of the age of Alice
=> Age of Jane = 80% of x = 80/100 * x = 0.8x
Adding the two ages gives 45.
The two ages are x and 0.8x.
The sum of the two ages is x+ 0.8 x = (1+0.8)x = 1.8x.
=> 1.8 x = 45
Dividing both sides by 1.8, x = 45/1.8
=> x= 25
But x represents the age of Alice.
So age of Alice is 25.
Age of Jane = 0.8* 25 = 20
Verification: Adding the two ages gives 45.
<span>given:
doris ran 2.5 KM
assume A=2500 M
she sprinted 300 M
B=300 M
she walked 1/4 distance she sprinted ie. 0.25*300=75 M
C=75 M
total= A+B+C
=2500+300+75=2875 M
so Doris travelled 2875 Meter during practice .</span>
Answer:
81
Step-by-step explanation:
Miguel is reading 3x the pages everyday. 1x3=3 3x3=9 9x3=27 27x3=81