Answer:
The time is less than or equal to 1.5 hours
Step-by-step explanation:
we know that
The speed is equal to divide the distance by the time
Let
x -----> the time in hours
The inequality that represent this situation is equal to
Solve for x
Multiply by x both sides
Divide by 70 both sides
Rewrite the inequality
therefore
The time is less than or equal to 1.5 hours
Answer: 70m - 34o
Step-by-step explanation:
4:
7:
9:
12:
y 5 10 15 20
A.
1
B.
2
C.
3
D.
5
Answer:
Option D - 5
Step-by-step explanation:
Given : Table
x 1 2 3 4
y 5 10 15 20
To find : The constant of variation for the relationship shown in the following table?
Solution :
The constant of variation means the relationship between variables does not change.
The constant of variation for an equation is , where k is the constant of variation.
Now, We have given the value of x and y substitute in the formula,
x=1,y=5
x=2,y=10
x=3,y=15
x=4,y=20
Therefore, The constant of variation is 5.
So, Option D is correct.
200 students would be reasonable to test because you can average the score more easily.
In plain English, an average is indeed a single number chosen to represent a group of numbers; it is often the sum of the values divided by the number of numbers in the group. The average of a integers 2, 3, 4, 7, or 9 is, for instance, 5. An average could be another statistic like the median or mode depending on the situation.
here, we have,
Because the mean would be greater if the personal incomes of a few billionaires were included, the median—the amount below which 50% of personal incomes fall and over which 50% of personal incomes rise—is sometimes used to represent the average personal income.
It is advised to avoid to use the word "average" while addressing central tendency measurements because of this.
If we wanted to know how high a 5th-grade student could jump, 200 students would be reasonable to test because you can average the score more easily.
Therefore, 200 students would be reasonable to test because you can average the score more easily.
To know more about average, here:
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