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Ierofanga [76]
3 years ago
7

Find the midpoint between two points on a number line if one of the points is at -7, and the other point is at 12.

Mathematics
2 answers:
vodomira [7]3 years ago
5 0

Step-by-step explanation:

\frac{12 - 7}{2}  = 2.5

quester [9]3 years ago
3 0

Answer:

D.  2.5.

Step-by-step explanation:

That would be (-7 + 12) / 2

= 5/2

= 2.5.

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Find the mean and median of the data. <br><br> 4.7,8.51,6.5,7.42,9.64,7.2,9.3
Murrr4er [49]

Answer:

Population size: 7

Median: 7.42

Minimum: 4.7

Maximum: 9.64

First quartile: 6.5

Third quartile: 9.3

Interquartile Range: 2.8

Outliers: none

5 0
3 years ago
Is this expression equivalent to 1.5?<br><br> 0.15 / 10<br><br> (0.15 divided by 10)
dimulka [17.4K]

Answer:

False

Step-by-step explanation:

0.15/10 equals 0.015

6 0
3 years ago
I need help please!!
guajiro [1.7K]

13-11 = 2

15-13 = 2

17-15 = 2

As you can see, the numbers increased by two. hence the answer is 2.

5 0
3 years ago
Endpoint: (9,-10), midpoint: (4,8)
Alika [10]

Answer:

( -1 , 26)

Step-by-step explanation:

So since you need to find the other endpoint, you would follow these steps:

1.) 4 = \frac{9 + x}{2\\}

2.) 8 = 9 + x  ( you just multiplied the 2 to the 4 to get 8)

3.) -1 = x  (just solve it like a regular equation, so just subtract 9 on both sides                  to get rid of it and that leaves you with -1 = x)

You took the x values of both points and put them in the equation.

And its the same for y

1.) 8 = \frac{-10 + y}{2}

2.) 16 = -10 + y

3.) 26 = y   (you added the 10 on both sides because the 10 was negative and that took the 10 out and so it left you with 26 = y)

5 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
4 years ago
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