13/15 of her allowance was spent in total.
To solve this, you'd need to find the least common denominator (LCD) so that both fractions have the same number on the bottom. In this case, the first number that you could get with 5 and 3 was 15.
Next, you'd have to multiply the numerator by the same amount as the denominator, so that the fractions are proportionate. So, for 1/5, since we had to multiply 5 by 3 to get 15, we'd multiply 1 by 3 as well, giving us 3/15. Doing the same with 2/3, we'd get 10/15.
Then, you add the two fractions together (10/15 + 3/15 = 13/15).
Now, in any other case, you could probably simplify the fraction after you've solved the problem. If we got 12/15 instead of 13/15, then we could simplify that to 4/5, since both 12 and 15 are divisible by 3. But in this case, this is the simplest form of that fraction.
Hope this helped!!!
Answer:
a. Class width=4
b.
Class midpoints
46.5
50.5
54.5
58.5
62.5
66.5
70.5
c.
Class boundaries
44.5-48.5
48.5-52.5
52.5-56.5
57.5-60.5
60.5-64.5
64.5-68.5
68.5-72.5
Step-by-step explanation:
There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,
Class
Interval frequency
45-48 1
49-52 3
53-56 5
57-60 11
61-64 7
65-68 7
69-72 1
a)
Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.
Class width=49-45=4
b)
The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.
Class
Interval Midpoints
45-48 
49-52 
53-56 
57-60 
61-64 
65-68 
69-72 
c)
Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.
Class
Interval Class boundary
45-48 44.5-48.5
49-52 48.5-52.5
53-56 52.5-56.5
57-60 56.5-60.5
61-64 60.5-64.5
65-68 64.5-68.5
69-72 68.5-72.5
Answer:
menos se repite es
9 porque despues del 4 5 6 7 8 ( 9 ) 10 y 11
We know that
if cos x is positive
and
sin x is negative
so
the angle x belong to the IV quadrant
cos x=5/13
we know that
sin²x+cos²x=1-------> sin²x=1-cos²x------> 1-(5/13)²---> 144/169
sin x=√(144/169)-------> sin x=12/13
but remember that x is on the IV quadrant
so
sin x=-12/13
Part A) <span>cos (x/2)
cos (x/2)=(+/-)</span>√[(1+cos x)/2]
cos (x/2)=(+/-)√[(1+5/13)/2]
cos (x/2)=(+/-)√[(18/13)/2]
cos (x/2)=(+/-)√[36/13]
cos (x/2)=(+/-)6/√13-------> cos (x/2)=(+/-)6√13/13
the angle (x/2) belong to the II quadrant
so
cos (x/2)=-6√√13/13
the answer Part A) is
cos (x/2)=-6√√13/13
Part B) sin (2x)
sin (2x)=2*sin x* cos x------> 2*[-12/13]*[5/13]----> -120/169
the answer Part B) is
sin(2x)=-120/169
By applying the definition of product between two <em>square</em> matrices, we find that ![\vec A \,\cdot \,\vec A = \left[\begin{array}{cc}1&2\\3&6\end{array}\right] \cdot \left[\begin{array}{cc}1&2\\3&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cvec%20A%20%5C%2C%5Ccdot%20%5C%2C%5Cvec%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%266%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%266%5Cend%7Barray%7D%5Cright%5D)
is equal to the matrix ![\vec A \,\cdot \,\vec A = \left[\begin{array}{cc}7&14\\21&42\end{array}\right]](https://tex.z-dn.net/?f=%5Cvec%20A%20%5C%2C%5Ccdot%20%5C%2C%5Cvec%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%2614%5C%5C21%2642%5Cend%7Barray%7D%5Cright%5D)
. (Correct choice: D)
<h3>What is the product of two square matrices</h3>
In this question we must use the definition of product between two <em>square</em> matrices to determine the resulting construction:
By applying the definition of product between two <em>square</em> matrices, we find that is equal to the matrix .
To learn more on matrices: brainly.com/question/11367104
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