How many significant figures does the number 546.000 have

1 answer:

3 0

3? Because a significant figure includes all digits except leading zeros or trailimg zeros. So if you take the zeros away you only have 3 numbers

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Solve the following system of equations algebraically:<br> y = x2 - 6x + 9<br> y = -9x + 19

Ratling [72]

{(-5,64), (2,1)}

linear equation: y = -9x + 19
quadratic equation: y = x² - 6x + 9

Substitute the y in the quadratic equation by the its value in the linear equation.

-9x + 19 = x² - 6x + 9
      - 19    - 19 *subtract 19 to both sides
-9x    = x² - 6x -10
+9x + 9x *add 9x to both sides
0 = x² + 3x - 10
0 = (x + 5) (x - 2) *Factor

Set each factor = 0 and solve
x + 5 = 0 ; x - 2 = 0
x = -5    ; x = 2

Find the corresponding value of y using the linear equation.
y = -9x + 19

x = -5 x = 2
y = -9(-5) + 19    y = -9(2) + 19
y = 45 + 19 y = -18 + 19
y = 64 y = 1

(-5,64) (2,1)

Check each value on each equation.
y = x² - 6x + 9
(-5,64) (2,1)
64 = (-5)² - 6(-5) + 9 1 = 2² - 6(2) + 9
64 = 25 + 30 + 9 1 = 4 - 12 + 9
64 = 64    1 = 1

y = -9x + 19
64 = -9(-5) + 19 1 = -9(2) + 19
64 = 45 + 19 1 = -18 + 19
64 = 64 1 = 1

{(-5,64), (2,1)}

3 0

3 years ago

Find the volume of the cylinder.

Georgia [21]

<h3>Answer:</h3><h3>Exact volume = 32pi</h3><h3>Approximate volume = 100.48</h3>

The approximate volume only applies when pi = 3.14

Use either answer, but not both of course.

===============================================

Work Shown:

V = volume of cylinder

V = pi*r^2*h

V = pi*2^2*8

V = pi*32

V = 32pi .... exact volume in terms of pi

V = 32*3.14

V = 100.48 .... approximate volume when we use pi = 3.14

6 0

3 years ago

The histogram shows the speeds of downhill skiers during a competition. Estimate the mean of the data set displayed in the histo

Ostrovityanka [42]

Answer:

≈50.6

Step-by-step explanation:

Not sure what precision level this problem is looking for, but for right-skewed distributions, we know that the mean is going to be pulled right and therefore the mean should be larger than the median. To a high confidence level, the mean should fall between 50 and 59, or in the third column.

If a single estimation is wanted, assume the values inside each column are uniformly distributed:

$\displaystyle\\\widehat{\mu}=\frac{34.5\cdot 4+...+74.5\cdot 2}{4+9+4+4+2}\approx \boxed{50.6}$