Answer:
Option A. 10 years
Step-by-step explanation:
we know that
Half-life is the amount of time required for a quantity to fall to half its value as measured at the beginning of the time period.
In this problem of a radioactive isotope is 10 years, which means that after 10 years half of the sample would have decayed and half would be left as it is.
After 10 years ( first half life) 5/2 = 2.5 g decays and 2.5 g remains left
Actually Welcome to the concept of Geometric Mean (leg) theorem.
So according to the rule, we know that, here altitude = h and left side is 6 and right side is 14 , so we get as,
==> altitude/left = right/altitude
==> h/6 = 14/h
==> h^2 = 14*6
==> h^2 = 84
2.) so now we apply the Pythagoras theorem in triangle ∆DCB
so we get as,
==> (14)^2 + h^2 = a^2
==> 196 + 84 = a^2
==> 280 = a^2
==> a = √280
==> after simplified => a = √70*4
==> a = 2√70 units
hence the correct answer is a => 2√70 units.
Answer:
1. (a) 32 (b) 8
2. (a) 48.33 (b) 92.75 (c) 9.63
3. We observe that sample mean of air quality index in Pomona is higher that that of Anaheim which indicates that the level of air pollution is higher in Pomona and there is more health concern here.
On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in air quality index in Pomona.
Step-by-step explanation:
First arranging our data of air quality index values for Pomona in ascending order, we get : 28, 42, 45, 48, 49, 50, 55, 58, 60.
Highest value in data - Lowest value in data = 60 - 28 = 32
(b) Interquartile rage = Third quartile - First quartile
=
=
observation in the data =
obs. { Because n = 9}
Therefore, = 42.
=
obs. = (3 x 2 ) =
obs = 50
So, Interquartile rage = 50 - 42 = 8.
2. (a) Sample mean formula = =
= 48.33
(b) Sample variance formula = where Xbar = Sample mean
= 92.75
(c) Sample standard deviation formula = =
= 9.63.
3. Given that a sample of air quality index readings for Anaheim has sample mean of 48.3, a sample variance of 136, and a sample standard deviation of 11.66.
We observe that sample mean of air quality index in Pomona is higher that that of Anaheim which indicates that the level of air pollution is higher in Pomona and there is more health concern here.
On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in air quality index in Pomona.