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Dominik [7]
3 years ago
10

A line graph titled Text messages sent has time of day on the x-axis and number of text on the y-axis. Girls sent 10 in the morn

ing, 25 in the afternoon, 40 in the evening, and 5 at night. Boys sent 5 in the morning, 15 in the afternoon, 20 in the evening, and 15 at night.
The graph shows the number of text messages sent by the boys and girls in one day. Complete the statements.

Boys sent more text messages than girls during the
.
In the morning, girls sent about
more text messages than boys.
Both boys and girls sent the most text messages in the
.
During the afternoon, the students sent a total of approximately
text messages.
Mathematics
2 answers:
slavikrds [6]3 years ago
7 0

Answer:

Boys sent more text messages than girls during the

✔ night

.

In the morning, girls sent about

✔ 5

more text messages than boys.

Both boys and girls sent the most text messages in the

✔ evening

.

During the afternoon, the students sent a total of approximately

✔ 40

text messages.

Step-by-step explanation:

klio [65]3 years ago
4 0

Answer:

Boys sent more text messages than girls during the

✔ night

.

In the morning, girls sent about

✔ 5

more text messages than boys.

Both boys and girls sent the most text messages in the

✔ evening

.

During the afternoon, the students sent a total of approximately

✔ 40

text messages.

Step-by-step explanation:

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Zanzabum
<h3>Answer:   0.5</h3>

This is equivalent to the fraction 1/2

==============================================================

Explanation:

The distance from A to B is 3 units. We can count out the spaces, or subtract the x coordinates of the two points and apply absolute value.

|A-B| = |-5-(-8)| = |-5+8| = |3| = 3

or

|B-A| = |-8-(-5)| = |-8+5| = |-3| = 3

We can say that segment AB is 3 units long.

--------------------------

The distance from A' to B' is 1.5 units because...

|A'-B'| = |-2.5-(-4)| = |-2.5+4| = |1.5| = 1.5

or

|B'-A'| = |-4-(-2.5)| = |-4+2.5| = |-1.5| = 1.5

The absolute values ensure the distance is never negative.

We can say A'B' = 1.5

---------------------------

Now divide the lengths of A'B' over AB to get the scale factor k

k = (A'B')/(AB)

k = (1.5)/(3)

k = 0.5

0.5 converts to the fraction 1/2.

The smaller rectangle A'B'C'D' has side lengths that are exactly 1/2 as long compared to the side lengths of ABCD.

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