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3 years ago

True or false? The first distribution shown below has a smaller mean than does the second.

Mathematics

1 answer:

cestrela7 [59]3 years ago

3 0

The awnser is It is true

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Parallel to y - 2x = 4 and passes through (0,5)

wolverine [178]

Answer:

y=2x+5

Step-by-step explanation:

y=2x+4 means that the slope is m1=2

A parallel line shall have the same slope so

m2=m1=2

Therefore, the equation of the new line will be

y=2x+q

where q is the y-intercept, that is 5 because the line must pass in (0,5), so

q=5

and

y=2x+5

4 0

3 years ago

Kong makes $14.25 per hour. If he works a 40-hour week, how much does he have to pay into FICA (7.65%)?

gizmo_the_mogwai [7]

Amount of FICA to be paid
($14.25 x 40) x (7.65/100) = $43.61

7 0

3 years ago

Solve 8(x+2)=36.......

brilliants [131]

We have to use the Distributive Property in this equation.
By the Distributive Property,
$8(x+2)=36
$
becomes
$8x+16=36.$
Now we need to isolate x. Remember PEMDAS?
$8x=20$
$x=20/8$
Simplify the right side.
$x=10/4$
$x=5/2$

3 0

3 years ago

Read 2 more answers

Find the value of x in the triangle

SVEN [57.7K]

Answer:

x = 112

Step-by-step explanation:

180 - 41 - 27 = 112

All the angles of a triangle need to add up to 180 degrees.

3 0

2 years ago

Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits

scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

$x^{2} + 3x - 8^{- 14} = 0$

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

The above solution is for the quadratic equation of the form:

$ax^{2} + bx + c = 0$

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

From the given eqn

a = 1

b = 3

c = $- 8^{- 14}$

Now, using the above values in the formula mentioned above:

$x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)$

Now, Rationalizing the above eqn:

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

$x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

Solving the above eqn:

$x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}$

Solving with the help of caculator:

$x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}$

The precise value upto three decimal places comes out to be:

$x_{1, 1'} = 0.758\times 10^{- 14}$

5 0

3 years ago