Answer: Stratified Random Sample
Step-by-step explanation:
The sample is a stratified random sample in the sense that each of the population that was sampled consisted of the necessary groups that need to be considered in their right proportion of representation which makes the sample valid for us to reasonably draw a conclusion of what to expect in the larger sample space. This sample favours every possible member of the population and helps us adequately plan for all
The fraction computed shows that the total quantity of meat that Joy bought is 5/6kg.
<h3>How to solve the fraction?</h3>
The fraction can be explained thus: Joy bought 1/2kg of meat from a store and bought 1/3kg of meat from store B. What is the total quantity of meat that Joy bought?
The total quantity of meat that Joy bought will be:
= 1/2 + 1/3
= 5/6
In conclusion, the correct option is 5/6kg.
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Answer: 8 hours
Step-by-step explanation: First, find the unit rate. For one hour of work, she earns 0.05263157894. Multiply this number 152.
Answer:
Step-by-step explanation:
Volume of tank is 3000L.
Mass of salt is 15kg
Input rate of water is 30L/min
dV/dt=30L/min
Let y(t) be the amount of salt at any time
Then,
dy/dt = input rate - output rate.
The input rate is zero since only water is added and not salt solution
Now, output rate.
Concentrate on of the salt in the tank at any time (t) is given as
Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000
dy/dt= dV/dt × dM/dV
dy/dt=30×y/3000
dy/dt=y/100
Applying variable separation to solve the ODE
1/y dy=0.01dt
Integrate both side
∫ 1/y dy = ∫ 0.01dt
In(y)= 0.01t + A, .A is constant
Take exponential of both side
y=exp(0.01t+A)
y=exp(0.01t)exp(A)
exp(A) is another constant let say C
y(t)=Cexp(0.01t)
The initial condition given
At t=0 y=15kg
15=Cexp(0)
Therefore, C=15
Then, the solution becomes
y(t) = 15exp(0.01t)
At any time that is the mass.
Answer:
Step-by-step explanation:
-3x + 4y = -18
8x - 4y = 28
5x = 10
x = 2
4 - y = 7
-y = 3
y = -3
(2, -3)