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Law Incorporation [45]

3 years ago

7

Here are the ages (in years) of 10 professors at a college. 44, 38, 45, 34, 28, 56, 54, 28, 61, 48 What is the percentage of the

se professors who are younger than 47?

1 answer:

olga55 [171]3 years ago

8 0

Well to be honest I don’t know that a tricky question

You might be interested in

The function f(x)is shown in the graph.<br><br> What is the equation for f(x)?

kirill115 [55]

Answer:

(-5/6)x - 1

Step-by-step explanation:

the equation is given by: y = mx + b

m = slope

b = y intercept

m = -5/6

y intercept = -1

f(x) = (-5/6)x - 1

8 0

3 years ago

Georgie needs to paint the wall with a fixing mixture.

leonid [27]

To determine if 3.5 liters of glue will be enough, you can set up a proportion using the ratio of glue to water (3:1).

<u>3 L glue </u> <u>3.5 L glue</u>
1 L water = x L water

Use cross products to solve for the amount of water.

<u>3x</u> = <u>3.5</u>
3 3

x = 1.17 L

You will need approximately 1.17 L of water with 3.5 L of glue. This is a total of 4.67 L, which is NOT enough.

You will <u>not</u> be able to make the mixture.

7 0

3 years ago

What is the x and y intercept of y=1/3x-2, y=12-6x, and y=3x-7? Please show work!

Lana71 [14]

Answer:

y = 1/3x - 2:

y - intercept: - 2
x - intercept: 6

y = 12 - 6x:

y - intercept: 12
x - intercept: 2

y = 3x - 7:

y - intercept: -7
x - intercept: 7/3

Step-by-step explanation:

Hello!

<h3>Slope Intercept Form</h3>

The slope-intercept form of a line is $y = mx + b$

y = output, ordinate
m = slope
x = input, abscissa
b = y-intercept

<h3>y = 1/3x - 2</h3>

The y-intercept is the "b" value of the line, given as -2.

To find the x-intercept, we simply replace the y-value with 0:

0 = 1/3x - 2
2 = 1/3x
6 = x

y - intercept: - 2

x - intercept: 6

____________________________________________________

<h3>y = 12 - 6x</h3>

You can rearrange the equation, y = -6x + 12, to find the y-intercept: 12.

Solving for the x-intercept (remember, replace y with 0):

0 = -6x + 12
-12 = -6x
2 = x

y - intercept: 12

x - intercept: 2

____________________________________________________

<h3></h3><h3>y = 3x - 7</h3>

Simply take the "b" value, -7, to find the y-intercept of -7.

Solve for the x-intercept:

0 = 3x - 7
7 = 3x
x = 7/3

y - intercept: -7

x - intercept: 7/3

<u><em>Additional Note:</em></u>

You can also solve for the y-intercept by replacing x with 0. This method can be used for any interests, just plug in the opposite variable from what you are solving for with 0.

The proof is that the x-intercept is the value when the y = 0, the x-axis. You can plug in 0 for y to solve for x. Similarly, when solving for the y-intercept, you plug 0 for x as the y-axis is at x = 0.

3 0

2 years ago

How are 25:5 and 10:2 equivalent ratios?

stira [4]

Answer:

Multipy by 2.5

Step-by-step explanation:

If you multiply 10 and 2 by 2.5 you would get 25 and 5

7 0

3 years ago

For what value of k are there two distinct real solutions to the original quadratic equation (k+1)x²+4kx+2=0.

lina2011 [118]

Answer:

k ∈ (-∞, $-\frac{1}{2}$ )∪(1,∞)

Step-by-step explanation:

For quadratic equations $ax^2+bx+c=0,a\neq 0$ you can find the solutions with the Bhaskara's Formula:

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\\and\\x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

A quadratic equation usually has two solutions.

If you only want real solutions the condition is that the discriminant ( $\Delta$ ) has to be greater than zero, this means:

$\Delta=b^2-4ac>0$

Then we have the expression:

$(k+1)x^2+4kx+2=0$

$a=(k+1)\\b=4k\\c=2\\$

Now to find two distinct real solutions to the original quadratic equation we have to calculate the discriminant:

$b^2-4ac>0\\(4k)^2-4.(k+1).2>0\\16k^2-8(k+1)>0\\16k^2-8k-8>0$

We got another quadratic function.

$16k^2-8k-8>0$ we can simplify the expression dividing both sides in 8.

$16k^2-8k-8>0\\\\\frac{16k^2}{8} -\frac{8k}{8} -\frac{8}{8} >\frac{0}{8}\\\\2k^2-k-1>0$

We can apply Bhaskara's Formula except that the condition in this case is that the solutions have to be greater than zero.

$2k^2-k-1>0\\a=2\\b=-1\\c=-1$

$k_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1+\sqrt{9} }{4}=\frac{1+3}{4} =1 \\and\\k_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}$

Then,

$k>1 \\and\\k$

The answer is:

For all the real values of k who belongs to the interval:

(-∞, $-\frac{1}{2}$ )∪(1,∞)

there are two distinct real solutions to the original quadratic equation $(k+1)x^2+4kx+2=0$

4 0

4 years ago