Could someone explain/answer these?
2 answers:
1)
f(x) = x² + 9 g(x) = 24 + 4x
f(4) = (4)² + 9 g(-1) = 24 + 4(-1)
= 16 + 9 = 24 - 4
= 25 = 20
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g(x) = 2ˣ - 4 f(x) = 4x² - 5x - 5
g(6) = 2⁶ - 4 f(-5) = 4(-5)² - 5(-5) - 5
= 64 - 4 = 100 + 25 - 5
= 60 = 120
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f(x) = (3x + 7)²
f(1) = (3(1) + 7)²
= (3 + 7)²
= (10)²
= 100
****************************************************************
3xy + 2x² - y³ ; when x = 6 and y = -2
3(6)(-2) + 2(6)² - (-2)³
-36 + 72 - (-8)
36 + 8
44
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Answer:
a: 1003 < µ < 1023
b: See attached photo for answers
c: see below
d and e: See second attached photo
f: See below
Step-by-step explanation:
a. to find a confidence interval, first find the error with the formula,
E = Z(σ/√n) where Z is the value of the z-score that relates to the confidence level, σ is the standard deviation of the population (or of the sample if the population standard deviation isn't given), and n is the sample size
For this problem, E = 1.96(108/√453) = 9.95 = 10
Now add and subtract that value form the sample average...
x - E < µ < x + E becomes 1013 - 10 < µ 1013 + 10
simplifies to 1003 < µ < 1023
b. See attached photo for hypothesis test results...
c. H0: µ = 1013
Ha: µ > 1013 (claim)
They want to test if it's greater than their first attempt, so this claim becomes the alternate hypothesis. We're testing against their first attempt, that's why µ = 1013, not 1000
d and e. See second attached photo for hypothesis test results...
f. There is enough evidence to support the claim that the students second attempt was an improvement from their first attempt. They happened to take a 2 hour tutoring course between tests. This doesn't prove that the tutoring was effective thought. It very well might have been effective, but also, the students taking the test again means they've already seen the test and know what to expect, which could have also boosted their scores.
