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gayaneshka [121]
3 years ago
15

Below is the data of ages at the local amusement park on a certain day.

Mathematics
1 answer:
nignag [31]3 years ago
6 0

Answer:

39

Step-by-step explanation:

To find the intervals you will need to find the lowest and highest numbers, in this case, it would be 1 and 35.  A general rule would be 5-7 intervals, I will use 5.

Here are the intervals with the number of people that were in each:

1-7 (9)

8-14 (14)

15-21 (8)

22-28 (3)

29-35 (5)

14+9+8+3+5=39

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Simplify -2+3(1-4)-2​
oksian1 [2.3K]

Answer:

-13

Step-by-step explanation:

−2+3(1−4)−2

=−2+(3)(−3)−2

=−2+−9−2

=−11−2

=−13

hope it helped

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4 0
3 years ago
The sides of a triangle are 7, 4, n. If n is an integer, state the largest and smallest possible values of n.
Sladkaya [172]

Answer:

4, 10

Step-by-step explanation:

The value for the third side of the triangle is given by

b-a < n < b+a where a and b are the two other sides of the triangle and b>a

7-4 < n < 7+4

3 < n < 11

Since n is an integer

4 would be the smallest value and 10 would be the largest

8 0
3 years ago
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Belinda runs 8 kilometers in 60 minutes. At this rate, how long would it take her to run 2 kilometers?
DiKsa [7]

Answer:

15 minutes

Step-by-step explanation:

4 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}
VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}

Rationalize the denominator by multiplying both numerator and denominator by √10:

\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}

8 0
3 years ago
Which statements are always true regarding the diagram? Check all that apply. m∠3 + m∠4 = 180° m∠2 + m∠4 + m∠6 = 180° m∠2 + m∠4
Nostrana [21]

Answer:

The true statements are:

m∠ 3 + m∠ 4 = 180° ⇒ 1st

m∠ 2 + m∠ 4 + m∠ 6 = 180° ⇒ 2nd

m∠ 2 + m∠ 4 = m∠ 5 ⇒ 3rd

Step-by-step explanation:

* Look to the attached diagram to answer the question

# m∠ 3 + m∠ 4 = 180°

∵ ∠ 3 and ∠ 4 formed a straight angle

∵ The measure of the straight angle is 180°

∴ m∠ 3 + m∠ 4 = 180° ⇒ <em>true</em>

# m∠ 2 + m∠ 4 + m∠ 6 = 180°

∵ ∠ 2 , ∠ 4 , ∠ 6 are the interior angles of the triangle

∵ The sum of the measures of interior angles of any Δ is 180°

∴ m∠ 2 + m∠ 4 + m∠ 6 = 180° ⇒ <em>true</em>

# m∠ 2 + m∠ 4 = m∠ 5

∵ In any Δ, the measure of the exterior angle at one vertex of the

  triangle equals the sum of the measures of the opposite interior

  angles of this vertex

∵ ∠ 5 is the exterior angle of the vertex of ∠ 6

∵ ∠2 and ∠ 4 are the opposite interior angles to ∠ 6

∴ m∠ 2 + m∠ 4 = m∠ 5 ⇒ <em>true </em>

# m∠1 + m∠2 = 90°

∵ ∠ 1 and ∠ 2 formed a straight angle

∵ The measure of the straight angle is 180°

∴ m∠1 + m∠2 = 90° ⇒ <em>Not true</em>

# m∠4 + m∠6 = m∠2

∵ ∠ 4 , ∠ 6 , ∠ 2 are the interior angles of a triangle

∵ There is no given about their measures

∴ We can not says that the sum of the measures of ∠ 4 and ∠ 6 is

  equal to the measure of ∠ 2

∴ m∠4 + m∠6 = m∠2 ⇒ <em>Not true</em>

<em></em>

# m∠2 + m∠6 = m∠5

∵ ∠ 5 is the exterior angle at the vertex of ∠ 6

∴ m∠ 2 + m∠ 6 = m∠ 5 ⇒ <em>Not true</em>

6 0
3 years ago
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