Answer:
Choice A: there is insufficient evidence to conclude that the mean for the fiber content is less than 2.5 gms.
Step-by-step explanation:
From the graphs given the more appropriate is one sample t- test
One sample t- test is used and the results are
Hypothesis Test Fiber
Sample Mean 2.62
Sample St. Dev 3.492
Degrees of Freedom 24
t- test statistic 0.1718
p- value 0.5675
the critical region for this test at ∝= 0.05 is t < -1.711
There is not enough evidence to reject the null hypothesis.
Since the calculated t= 0.1718 does not fall in the critical region the null hypothesis is accepted that the mean fiber content is not less than 2.5 grams.T
Slope is equal to the change in
y
over the change in
x
, or rise over run.
m = change in y / change in x
The change in
x
is equal to the difference in x-coordinates (also called run), and the change in
y
is equal to the difference in y-coordinates (also called rise).
m = y^2 - y^1 / x^2 x^1
Substitute in the values of
x
and
y
into the equation to find the slope.
m
=
3
−
(
−
6
)
8
−
(
−
15
)
Simplify.
Simplify the numerator.
Tap for more steps...
m
=
9
8
−
(
−
15
)
Simplify the denominator.
Multiply
−
1
by
−
15
.
m
=
9
8
+
15
Add
8
and
15
.
m
=
9/23
Answer:
To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.
Assume θ with a value and substitute with it.
Let θ = 45°
So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2
R.H.S = cot θ = cot 45 = 1
So, L.H.S ≠ R.H.S
So, sin θ cos θ = cot θ is not a trigonometric identity.
Answer:
He set the compass on a, with the width slightly wider than the distance to o and drew an arc above and below o. He then set the compass on b, without changing the width on the compass and did the same thing. True
He drew a line through where the arc pairs intersected and then labeled the points where this line crossed the top and bottom of the original circle. True
He set the compass on o, with the width slightly wider than the distance to b, and he drew a slightly larger circle around the original circle. False
He used the points that were marked along the circumference of the original circle as the vertices for the square. True
Step-by-step explanation:
Performance matters test perhaps?
Answer:
b
Step-by-step explanation:
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