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Mkey [24]
2 years ago
11

What is the sum of the 4th square number and the 12th square number

Mathematics
1 answer:
givi [52]2 years ago
7 0

Answer:

16

Step-by-step explanation:

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What is the range of the data in this stem-and-leaf plot?
sergij07 [2.7K]
The range is the difference between the lowest and the highest number.

the answer is 30.5
3 0
3 years ago
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WILL GIVE BRAINLIEST IF CORRECT
SCORPION-xisa [38]

9514 1404 393

Answer:

  5544 square units

Step-by-step explanation:

Translation does not change the dimensions of the figure. The two figures together tell you the right triangle has a base of 154 units and a height of 72 units.

The area of a triangle is given by ...

  A = 1/2bh

  A = 1/2(154 units)(72 units) = 5544 square units

The areas of each of the triangles are 5544 square units.

8 0
2 years ago
heather has divided $6300 between two invesments, one paying 9%, the other paying 4%. If the return on her investment is $372, h
Ganezh [65]
Assuming that the two investments are  X & Y
X + Y = 6300
X = 6300 - Y                                       (1)

9/100X + 4/100 Y = 372                       (2)
replacing X from (1) into (2)
9/100(6300-Y) + 4/100 Y = 372
567 - 9/100Y +4/100Y = 372
(-9+4)/100Y = 372 - 567
5/100Y = -195
Y = 100*195/5 = 3900
From (1) we can get X
X = 6300 - 3900 = 2400

I hope this is helpful 



6 0
3 years ago
Name the intersection of plane acg and plane bcg
cestrela7 [59]
Please attach a photo so I may help you.
8 0
3 years ago
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If you are dealt 4 cards from a shuffled deck of 52 cards, find the probability of getting two queens and two kings.
Bingel [31]

<u>Given</u>:

If you are dealt 4 cards from a shuffled deck of 52 cards.

We need to determine the probability of getting two queens and two kings.

<u>Probability of getting two queens and two kings:</u>

The number of ways of getting two queens is 4C_2

The number of ways of getting two kings is 4C_2

Total number of cases is 52C_4

The probability of getting two queens and two kings is given by

\text {probability}=\frac{\text {No.of fanourable cases}}{\text {Total no.of cases}}

Substituting the values, we get;

probability=\frac{4C_2 \cdot 4C_2}{52C_4}

Simplifying, we get;

probability=\frac{6 (6)}{270725}

probability=\frac{36}{270725}

probability=0.000133

Thus, the probability of getting two queens and two kings is 0.000133

7 0
3 years ago
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