Answer:
-8/45
Step-by-step explanation:
we can solve a/b by plugging in the given values and we have the new expression:
4/5 / -9/2
to solve, we can use multiply the two fractions together using KCF which stands for: Keep Change Flip. we keep the first fraction as is, change the division sign into a multiplication, and flip the 2nd fraction to its reciprocal (numerator becomes denominator and denominator becomes numerator)
4/5 / -9/2 becomes:
4/5 × 2/-9 < simply multiply the fraction
4 × 2 = 8
5 × -9 = -45
8/-45 is our new fraction. we cannot simplify this any further so this is our answer which can also be written as -8/45
Answer: M = - 2/3
Step-by-step explanation:
Not sure this should be the answer tho
Answer:
7.5
Step-by-step explanation:
To solve this expression, we can use PEMDAS or order of operations.
<u>Parenthesis</u>
<u>Exponents</u>
<u>Multiplication>Division (Depends on which comes first from left to right)</u>
<u>Addition>Subtraction (See Multiplication>Division)</u>
So First, lets solve the <u>Parenthesis</u>. -2/5=-.4
So now, our equation is -3/-.4
There is no exponents or multiplication, so lets just <u>Divide</u>.
-3/-.4=7.5
The solution is 7.5
Hope this helps!
Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
Answer:
Option 4, 64 cm^2
Step-by-step explanation:
The polygon in the shape of a star can be made to be easier to find the area of by cutting it into triangle. For all the "point edges" of the star, we can cut them in half from the vertex to create two right-angled triangles at each edge.
The formula of the area of a triangle is bh/2: the height is given as 6cm, the base is 4cm /2 (after you cut it in half) and so is 2 cm. Since the edges can be cut into two, there are 8 right angled triangles at the edges in total.
To find the area of them:
A=8(bh/2)=8(2*6/2)=48 cm^2
However, we are still missing the part at the centre we haven't found the area of, the square. The area of a square is given by the square of the side length(4cm). Thus, A=4*4=16cm^2
Adding the areas of all the edges and the square:
48 + 16 = 64cm^2
Hence, the area of the polygon is 64cm^2
