Solution: The missing reason in Step 8 is substitution of 
.
Explanation:
The given steps are used to prove the formula for law of cosines.
From step 5 it is noticed that our equation is
..... (1)
From step 7 it is noticed that the value of 
is 
.
So by substituting for in equation (1) we get the equation of step 8, i.e.,
Hence, the missing reason in Step 8 is substitution of .
the only statement that is true is the one in option D.
"Bar graphs are used to represent data that is discrete".
<h3>Which of the statements are true regarding dot plots, bar graphs, and histograms?</h3>
Dot plots, bar graphs, and histograms are used to repersent graphically data sets.
Thus, what these graphs do represent are populations in a data set with a given property.
Remember that data is usually discrete, so we usually use dot plots and bar graphs to represent discrete data.
Histograms show distributions of numerical data (it can be used for continuous or discrete data).
With all that in mind, we conclude that the only statement that is true is the one in option D.
"Bar graphs are used to represent data that is discrete".
If you want to learn more about data sets:
brainly.com/question/3514929
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Answer:
the answer is 1.2370 .. give me brainliest
Answer:
We can reject the hypothesis at the 0.05 significance level.
Step-by-step explanation:
given that an urn contains a very large number of marbles of four different colors: red, orange, yellow, and green. A sample of 12 marbles drawn at random from the urn revealed 2 red, 5 orange, 4 yellow, and 1 green marble.
H0: All colours are equally likely
Ha: atleast two colours are not equally likely
(Two tailed chi square test at 5% significance level)
color Red Orange Yellow Green total
Observed O 2 5 4 1 12
Expected E 3 3 3 3 12
(O-E)^2/E 0.3333 5.3333 0.3333 5.3333 11.3333
chi square = 11.3333
df =3
p value = 0.010055
Since p < 0.05 we reject H0
We can reject the hypothesis at the 0.05 significance level.
9514 1404 393
Answer:
216.66 yd²
Step-by-step explanation:
One way to find the area of a ring like this is to multiply its centerline length by the width of the ring.
Here, the diameter of the circle that is the centerline of the ring is 23 yd. The circumference of that circle is ...
C = πd = 3.14(23 yd) = 72.22 yd
Then the ring area is ...
(72.22 yd)(3 yd) = 216.66 yd² . . . area of shaded region
