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Alja [10]
1 year ago
13

Why is the slope of line segment AC the same as the slope of line segment DF in the figure below?

Mathematics
2 answers:
AlekseyPX1 year ago
8 0

Answer:

Answer A because AB/EF=BC/DE?

Step-by-step explanation:

This is due to the properties of similar triangles ABD and FED where ratio of corresponding sides between similar triangles are constant.

Wewaii [24]1 year ago
3 0

Answer:

The answer would be A. because AB/EF=BC/DE

Step-by-step explanation:

AB = 2

EF = 5

BC = 1

DE = 2.5

2/5 = 0.4

1/2.5 = 0.4

The equation AB/EF=BC/DE is correct. All of the others are not.

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Theo recorded the means and mean absolute deviations of his language arts and Biology scores. He found the difference in the mea
Rina8888 [55]

The answer is actually A.2 because I just took the test.

8 0
2 years ago
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Susan is attending a talk at her son's school. There are 8 rows of 10chairs where 54 parents are sitting. Susan notices that eve
kozerog [31]

Answer:

The largest possible number of adjacent empty chairs in a single row is 3

Step-by-step explanation:

The parameters given are;

The number of chairs = 8 × 10 = 80 chairs

The number of parents = 54

Sitting arrangements of parents = Alone or to one other person

Therefore;

The maximum number of parents on a row = 1 + 1 + 0 + 1 + 1 + 0 + 1 + 1 + 0 + 1 = 7

Hence when the rows have the maximum number of parents occupying the seats we have for the 8 rows;

7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 56

But there are only 54 parents, therefore, up to the 7th row will have 7 parents while the 8th row will have only 5 parents to make the possible sitting arrangement to be as follows;

7 + 7 + 7 + 7 + 7 + 7 + 7 + 5 = 54

The sitting arrangement for the 8th row is therefore

1 + 1 + 0 + 1 + 1 + 0 + 1 + 0 + 0 + 0

Hence there will be three empty seats in the 8th row making the largest possible number of adjacent empty chairs in a single row = 3.

4 0
2 years ago
One morning, eight buses arrive at a bus stop.
Semenov [28]

Answer:

0 and 5 minutes

Step-by-step explanation:

Median is the middle observation of given data. It can be found by following steps:

Arranging data in ascending or descending order.

Taking the average of middle two value if the total number of observation is even, and this average is our median.

or, if we odd number of observation then the most middle value is our median.

The mode is the observation which has a high number of repetitions (frequency).

We have given 8 numbers and we have to find the other two numbers a minute late for each bus to arrive at the bus stop.

Also, we have given the median and mode of all 10 buses i.e. 3.5 and 0 respectively.

Since the mode is 0, so the frequency of 0 must be greater than 2 times and less than or equal to 4 times.

The frequency of 0 is 4 is not possible as then we didn't get a median 3.5 then.

So the frequency of 0 is 3.

Hence one of number is 0 in two unknown number of a minute late for each bus arrive at the bus stop.

Now, we have the number of minutes late for each bus after arranging them in ascending order is: 0 0 0 2 2 6 8 8 9

For getting a median 3.5 we must have a number between 2 and 6. Let it be x then:

\frac{2+x}{2} = 3.5

⇒x = 5

Hence the number of minutes late for each bus is: 0 0 0 2 2 5 6 8 8 9

Thus, 0 and 5 are minutes late for the two-afternoon buses.

6 0
2 years ago
Read 2 more answers
Factor 10e2 completely.
Reil [10]

Answer:

(2•5e2)

Step-by-step explanation:

3 0
1 year ago
1. The population of a town was 5655 in 2010. The population grows at a rate of 1.4% annually.
DaniilM [7]

Answer:

Step-by-step explanation:

a) We would use the formula for exponential growth model which is expressed as

y = b(1 + r)^t

Where

y represents the population after t years.

t represents the number of years.

b represents the initial population.

r represents rate of growth.

From the information given,

b = 5655

r = 1.4% = 1.4/100 = 0.014

Therefore, the equation would be

y = 5655(1 + 0.014)^t

y = 5655(1.014)^t

b) in 2022, t = 2022 - 2010 = 12 years

y = 5655(1.014)^12

y = 6682

6 0
2 years ago
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