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Paul [167]
4 years ago
5

Round 45,809 to nearest thousand and also round 26,789 to nearest ten thousand

Mathematics
2 answers:
julia-pushkina [17]4 years ago
7 0
45,809 rounded to the nearest thousand is 46,000.

26,789 rounded to the nearest ten thousand is 30,000
Yuki888 [10]4 years ago
4 0
The nearest thousand would be the number to the left of the comma. So in this case it is the 5, this means that if it is 5 or more we raise the 5 to a 6 and make everything behind it a 0 or if it is 4 or less we keep it at 5 and make everything behind it 0. 
So it would be 46,000 since it is an 8 and the next one would be 27,000 because the next number is 7.
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EastWind [94]

Answer:

x=\frac{-363}{70} ≈ -5.1857

y=\frac{-192}{35} ≈ -5.4857

z=\frac{313}{84} ≈ 3.7262

Step-by-step explanation:

Rewrite the equation system as:

6x+8y=-75

-3x+6y+6z=5

2x-9y=39

Now, write the system in its augmented matrix form:

\left[\begin{array}{cccc}6&8&0&-75\\-3&6&6&5\\2&-9&0&39\end{array}\right]

applying row reduction process to  its associated augmented matrix:

Swap R1 and R3, and then Swap R1 and R2:

\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\6&8&0&-75\end{array}\right]

R3+2R1

\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\0&20&12&-65\end{array}\right]

3R2+2R1

\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&20&12&-65\end{array}\right]

15R3+20R2

\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&0&420&1565\end{array}\right]

Now we have a simplified system:

-3x+6y+6z=5\\0-15y+12z=127\\0+0+420z=1565

-3x+6y+6z=5\hspace{5 mm}(1)\\0-15y+12z=127\hspace{3 mm}(2)\\0+0+420z=1565\hspace{3 mm}(3)

From (3):

z=\frac{313}{84} (4)

Replacing (4) in (2)

y=\frac{-192}{35} (5)

Finally replacing (5) and (4) in (1)

x=\frac{-363}{70}

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3 years ago
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Airida [17]

Answer:

7 6/18

Step-by-step explanation:

8 0
3 years ago
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Pavlova-9 [17]
Since A (3,4) A’(9,12)
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3 0
3 years ago
6.1.3
Nata [24]

Answer:

The requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.

Step-by-step explanation:

A normal-distribution is an accurate symmetric-distribution of experimental data-values.  

If we create a histogram on data-values that are normally distributed, the figure of columns form a symmetrical bell shape.  

If X \sim N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z \sim N (0, 1).

The distribution of these z-variates is known as the standard normal distribution.

Thus, the requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.

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