Pleaaasseee help!!! BRAINLIEST FOR CORRECT ANSWER!! What is the transformation rule for a translation 3 units right?
1 answer:
Answer:
Rule: replace x by x - a where a is the number of units that you want to move right. a must be greater than 0. x - - a would move left.
Step-by-step explanation:
You want f(x) to move 3 units to the right.
That would mean that x would be replaced by x - 3. Just to be sure let's try it.
- Suppose you have f(x) = x^2 + 6x + 5 It is graphed as the red line
- Now suppose you want to move 3 units right.
- It would replaced like f(x - 3) = (x - 3)^2 + 6(x - 3) + 5 which is the blue line
- Notice nothing else is changed. The blue line looks exactly like the red line except that it is shifted 3 units to the right.
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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!