If ΔABC has vertices A (3, –1), B (2, 1), and C (5, 3), find vertices of its image ΔA′B′C′ after dilation with a scale factor of
1 answer:
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Answer:
Step-by-step explanation:
Step 1: Define
Difference Quotient:
f(x) = -x² - 3x + 1
f(x + h) means that x = (x + h)
f(x) is just the normal function
Step 2: Find difference quotient
- <u>Substitute:</u>
- <u>Expand and Distribute:</u>
- <u>Distribute:</u>
- <u>Combine like terms:</u>
- <u>Factor out </u><em><u>h</u></em><u>:</u>
- <u>Simplify:</u>
Answer:
a) A sample of 256 was used in this survey.
b) 45.14% probability that the point estimate was within ±15 of the population mean
Step-by-step explanation:
This question is solved using the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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.
a. How large was the sample used in this survey?
We have that . We want to find n, so:
A sample of 256 was used in this survey.
b. What is the probability that the point estimate was within ±15 of the population mean?
15 is the bounds with want, 25 is the standard error. So
Z = 15/25 = 0.6 has a pvalue of 0.7257
Z = -15/25 = -0.6 has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
45.14% probability that the point estimate was within ±15 of the population mean