12 and 3/10 more than 5 and 13/1000 of d equals 15 and 302/1000
12 and 3/10+(5 and 13/1000 times d)=15 and 302/1000
convert to improper fractions
12 and 3/10=123/10
5 and 13/1000=5013/1000
15 and 302/1000=15302/1000
123/10+(5013/1000 times d)=15302/1000
subtract 123/10 from both sides
123/10=12300/1000
(15302-123000)/1000=2698/1000
5013/1000 times d=2698/1000
multiply both sides by 1000/5013 to clear fraction
d=2698/5013
Answer: a) Yes b) Polygon ABCD ~ Polygon EFGH c) JKLM is 3x bigger than polygon ABCD
Step-by-step explanation:
(for part c)
JM= 36
AD= 12
36/12= 3
JKLM is 3x bigger than polygon ABCD
Answer:
Group b most likely has a lower mean age of salsa students
Step-by-step explanation:
Arithmetic Mean of the data is the average of a set of numerical values, calculated by adding them together and dividing by the number of terms in the set.
Here we are given with two groups that are Group A and Group B
both having total number of students = 20
Here the mean age of the data is addition of the all ages of different students divided by total number of students.
For group a
total age of the group = 3 × 5 + 4 × 10 + 6 × 17 + 4 × 24 + 3 × 29
= 15 + 40 + 102 + 96 + 87
=340
The mean age of salsa students= 340 ÷ 20 = 17
For group b
total age of the group = 6 × 7 + 3 × 10 + 4 × 14 + 5 × 16 + 2 × 21
= 42 + 30 + 56 + 80 + 42
=250
The mean age of salsa students= 250 ÷ 20 = 12.5
So the group b most likely has a lower mean age of salsa students
Learn more about Arithmetic Mean here - brainly.com/question/24688366
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Answer:
C
Step-by-step explanation:
You can reflect point x across the origin to find is is just above -1. Adding 3 to that value gives a point just above -1+3 = 2. That's where point C is located.
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<em>Algebraic solution</em>
Algebraically, you can set y = -x +3, and solve for x. That gives you ...
x = 3 -y
Using the relation given at the start:
0 < x < 1
0 < 3 -y < 1 . . . . . substitute for x
Adding y gives ...
y < 3 < y+1
We can separate this into two inequalities:
y < 3, and
3 < y+1
2 < y . . . subtract 1
Now, we have ...
2 < y < 3 . . . . . the location of point C
As you can see, it is much easier to use the number line directly to find the desired point.
Answer:
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