(HELP NEEDED ASAP!!!)The sequence below represents Marisa’s fine at the library for each day that she has an overdue book: $0.50
2 answers:
Answer:
C.
Step-by-step explanation:
We have been given a sequence representing the Marisa's fine at the library for each day that she has an overdue book: $0.50, $0.65, $0.80, $0.95, $1.10, ...
An arithmetic sequence is in form , where
= nth term of sequence.
= 1st term of the sequence.
d= Common difference.
We can see from our sequence that 1st term of our sequence is 0.50.
Let us find common difference by subtracting 0.50 from 0.65.
Upon substituting our values in arithmetic sequence we will get,
Upon distributing 0.15 we will get,
Now let us combine like terms.
Therefore, the equation will represent Marisa’s library fine as a function of a book that is n days overdue and option C is the correct choice.
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Answer:
We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.
Step-by-step explanation:
We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.
From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.
<u><em>Let </em></u><u><em> = average miles for deluxe tires</em></u>
So, Null Hypothesis, :
50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}
Alternate Hypothesis, :
< 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}
The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean lifespan = 46,500 miles
= population standard deviation = 8000 miles
n = sample of tires = 28
So, <u><em>test statistics</em></u> =
= -2.315
The value of the test statistics is -2.315.
Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.
Answer:
X ~ B(n, p)
Given: X ~ B(2, 5/17)
⇒ n = 2
⇒ p = 5/17
Binomial distribution formula:
As n = 2, calculate P(X = 0), P(X = 1) and P(X = 2):
**Probability Distribution Table attached**
