What is 6x+7y=4x+4y6x+7y=4x+4y
1 answer:
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Check the one-sided limits:
If <em>f(x)</em> is to be continuous at <em>x</em> = 5, then these two limits should have the same value, which means
5<em>k</em> = 200
<em>k</em> = 200/5
<em>k</em> = 40
Answer:
96.49% probability that the mean rent is more than $1100
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
What is the probability that the mean rent is more than $1100?
This is 1 subtracted by the pvalue of Z when X = 1100. So
By the Central Limit Theorem
has a pvalue of 0.0351
1 - 0.0351 = 0.9649
96.49% probability that the mean rent is more than $1100
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!