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2 years ago

Round 16.748 to the ones place

Mathematics

2 answers:

IceJOKER [234]2 years ago

4 0

17 for ones place.. 16.7 for tenths place.. 16.75 for hundredths place and so on..

Aleonysh [2.5K]2 years ago

3 0

So , what helps me in math is my teachers was us say 4 or lest let it rest 5 or more let it sore so what dose the 8 tell the 4 USE A ARROW if needed. thank you

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Please thank youuu asap!!!!!!!!!!!!!!!!!

Vitek1552 [10]

All direct proportional graphs go through the origin (0,0). So this is where one point could go. The second point is found by going up 5 units and then to the right 3 units to get to (3,5)

<h3>Draw a straight line through (0,0) and (3,5) to graph</h3>

The equation of this line is y = (5/3)x

5 0

2 years ago

The heights of women aged 20 to 29 are approximately normal with mean 64 inches and standard deviation 2.7 inches. Men the same

san4es73 [151]

Answer:

Step-by-step explanation:

$Z=\frac{x-mu}{\sigma }$

X - Heights of women aged 20 to 29: X is N(64, 2.7)

Y - Heights of men aged 20 to 29: X is N(69.3, 2.8)

a) X=6'=72"

Z= $\frac{72-64}{2.7} =2.963$

b) Y = 72"

z= $\frac{72-69.3}{2.8} =0.965$

c) 72' women can be taken as very tall but men moderately taller than average

d) Y<5'5" =65"

Z< $\frac{65-69.3}{2.8} =-1.536$

e) X>73"

Z> $\frac{73-64}{2.7} =3.33$

Almost 0% can be taken

f) IQR =Q3-Q1

Z scores =

5 0

3 years ago

Rewrite the system of linear equations as a matrix equation AX = B.

iren2701 [21]

Answer:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

7 0

3 years ago

You start at (2, -4). You move up 4 units. Where do you end?

NeTakaya

Answer:

(2, 0)

Step-by-step explanation:

Hope This Helps! : )

7 0

2 years ago

What are corresponding angles<br>

Ratling [72]

Answer:

Corresponding angles are equal. They're a F shape and it could be upside down.

8 0

1 year ago