Answer:
13.4
Hope it helps that is what I think not that sure but I thing is corrwct!
Answer: -4/1 = 1/-4
= 1/-4
Just flip the fraction:)
Step-by-step explanation:
Answer:
we will fail to reject the null hypothesis and conclude that the mean pressure is not different from 4.7 psi
Step-by-step explanation:
Let's first define the hypothesis;
Null hypothesis: H0: μ = 4.7
Alternative hypothesis: Ha: μ ≠ 4.7
We have;
Sample size; n = 110
Sample mean; x¯ = 4.6
Variance: σ² = 0.64
Standard deviation; σ = √0.64 = 0.8
Formula for test statistic is;
z = (x¯ - μ)/(σ/√n)
z = (4.6 - 4.7)/(0.8/√110)
z = -0.1/0.0763
z = -1.31
From online p-value from z-score calculator attached, using; z = -1.31, two tailed hypothesis, significance value of 0.1, we have;
P-value = 0.190196
The p-value is greater than the significance value and thus we will fail to reject the null hypothesis and conclude that the mean pressure is not different from 4.7
σ μ
Answer:
3 1/5 ounces.
Step-by-step explanation:
That would be 16/5
= 3 1/5 ounces.
Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
