I will give brainliest to the best answer to this question: Each year, the Department of Water Works measures the chloride conce
2 answers:
Answer:
The total change that will likely occur in 2 years is mg/L
Step-by-step explanation:
Given : The first year, they found the chloride concentration changed by mg/L .
It is estimated that the chloride concentration will change mg/L the next year.
To find : What is the total change that will likely occur in 2 years?
Solution :
The chloride concentration changed by mg/L in first year.
The chloride concentration will change mg/L the next year.
The total change that will likely occur in 2 years is
In a mixed fraction - mg/L
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Answer:
The correct answer is "(18, 30) minutes".
Step-by-step explanation:
Mean,
= 24
Standard deviation,
= 2
The range of minutes where 99.7% of individuals completed the quiz will be:
=
On substituting the values, we get
=
=
now,
=
=
Or,
=
=
i.e., 18 to 30 minutes
Answer:
statistic value t= 5.43
p-value: < 0.00001.
Step-by-step explanation:
Hello!
To obtain information over the corrosion-resistance properties of a certain type of steel conduit a random sample of 45 specimens was taken and buried for two years.
The study variable is:
X: Max. penetration of a steel conduit.
The data of the sample
n= 45
sample mean X[bar]= 53.4
sample standard deviation S= 4.2
The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50
The hypothesis is:
H₀: μ ≤ 50
H₁: μ > 50
α: 0.05
To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.
Under the null hypothesis, the critical region and the p-value are one-tailed.
Critical value:
Rejection rule:
Reject the null hypothesis when t ≥ 1.68
t=
t= 5.43
The calculated value is greater than the critical value, thedecision is to rject the null hypothesis.
p-value:
P(t ≥ 5.43) = 1 - P(t < 5.43) = < 0.00001.
The p-value is less than α so the decision is to reject the null hypothesis.
Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.
I hope it helps!