At a high school basketball game the Lions and the Eagles are playing. The Lions attempted 18 free throws and made 10, attempted
2 answers:
Answer:
(a) 55.6%
(b) 45.0%
(c) 38.5%
(d) 27.1%
(e) 64
(f) 47
(g) The Lions
Step-by-step explanation:
(a) Free throw % for Lions is
(b) Free throw % for Eagles is
(c) For the Lions, they attempted a total of 51 + 14 = 65 field goals and made 21 + 4 = 25.
Field goal % for Lions is
(d) For the Eagles, they attempted a total of 45 + 14 = 59 field goals and made 10 + 6 = 16.
Field goal % for Eagles is
(e) Total points by Lions = (10 × 1) + (21 × 2) + (4 × 3) = 10 + 42 + 12 = 64
(f) Total points by Eagles = (9 × 1) + (10 × 2) + (6 × 3) = 9 + 20 + 18 = 47
(g) The Lions won because they had more points.
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Answer:
Step-by-step explanation:
First we need to make sure that the leading coefficient is a 1. Ours is a 4, so we need to factor it out, leaving us with
To complete the square, take half the linear term, square it, then add it to both sides. But don't forget about that 4 hanging around out front, refusing to be ignored. Our linear term is 18. Half of 18 is 9, and 9 squared is 81. Add 81 into the parenthesis, but what we REALLY added in was 4*81 which is 324:
To solve this, we need to get the x terms all by themselves. So let's divide both sides by 4 to get
The process of completing the square created a perfect square binomial on the left. We will state this binomial now:
We isolate the x term by taking the square root of both sides:
x - 9 = ±9
From that we have 2 equations:
x - 9 = 9 and x - 9 = -9
Which means that x = 18 or x = 0
Answer:
There is not sufficient evidence to support the claim that the technique performs differently than the traditional method.
Step-by-step explanation:
The null hypothesis is:
The alternate hypotesis is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
A researcher used the technique with 260 students and observed that they had a mean of 94 hours. Assume the standard deviation is known to be 6.
This means, respectively, that
The test-statistic is:
The pvalue is:
2(P(Z < -2.69))
P(Z < -2.69) is the pvalue of Z when X = -2.69, which looking at the z-table, is 0.0036
2*(0.0036) = 0.0072
0.0072 < 0.01, which means that the null hypothesis is accepted, that is, there is not sufficient evidence to support the claim that the technique performs differently than the traditional method.