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BaLLatris [955]
3 years ago
13

Find the value of 5 ∙ 23. 30 40 1,000 ASAP

Mathematics
2 answers:
True [87]3 years ago
5 0

The value of 5 ∙ 23 is 115.

The small dot is typically used as a multiplication symbol. When we multiply these two numbers the result is 115.

Shkiper50 [21]3 years ago
4 0

Answer:

mr grumby your slow ..... the 3 is the above the 2 not beside it , try again DUB

Step-by-step explanation:

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What is the greatest number that rounds to 500 when rounded to the nearest ten?
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I believe the answer is 549.
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The Thomas Supply Company Inc. is a distributor of gas-powered generators. As with any business, the length of time customers ta
Dmitry [639]

Answer:

The answer is below

Step-by-step explanation:

When computing quartile and decile, the data must be arranged in ascending order.

Given the data points:

13 13 13 20 26 26 29 31 34 34 35 35 36 37 38  41 41 41 42 43 46 47 48 49 53 55 56 62 67 82

The numbers are arranged in ascending order. The total number of terms is 30.

a)

First \ quartile(Q_1)=\frac{1}{4}(n+1)^{th}\  term\\\\where\ n=number\ of\ terms. Hence:\\\\Q_1=\frac{1}{4}(30+1)=7.75^{th}\ term=average\ of\ 7th\ and\ 8th\ term=\frac{29+31}{2} =30\\\\Q_1=30\\\\Third \ quartile(Q_3)=\frac{3}{4}(n+1)^{th}\  term\\\\Q_3=\frac{3}{4}(30+1)=23.25^{th}\ term=average\ of\ 23rd\ and\ 24th\ term=\frac{48+49}{2} =48.5\\\\Q_3=48.5

b)

Second \ decile(D_2)=\frac{2}{10}(n+1)^{th}\  term\\\\where\ n=number\ of\ terms. Hence:\\\\D_2=\frac{2}{10}(30+1)=6.2^{th}\ term=average\ of\ 6th\ and\ 7th\ term=\frac{26+29}{2} =27.5\\\\D_2=27.5\\\\\\Eight \ decile(D_8)=\frac{8}{10}(n+1)^{th}\  term=\frac{8}{10}(30+1)=24.8^{th}\ term\\=average\ of\ 24th\ and\ 25th\ term=\frac{49+53}{2} =51\\\\D_8=51

6 0
3 years ago
The question is in the picture. (Alegbra 1)
morpeh [17]

Answer:

Step-by-step explanation:

4/5a*3/2a^2

4*3/5a*2a^2

12/10a^3

6/5a^3

therefore ur answer is 6/5a^3

6 0
3 years ago
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tangare [24]

Answer:

  (3)  y = 4x

Step-by-step explanation:

In order for the equation not to change, the point (0, 0) must be on the original line and so on the line after dilation. The only equation with (0, 0) as a point on the line is y=4x.

Dilation about the origin moves all points away from the origin some multiple of their distance from the origin. If a point is on the origin, it doesn't move. We call that point the "invariant" point of the transformation. For the equation of the line not to change, the invariant point must be on the line to start with.

8 0
3 years ago
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