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3 years ago

Please help! i need to find the mean

Mathematics

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

Step-by-step explanation:

$\overline x =\dfrac{-3\cdot4+(-2)\cdot2+3\cdot1+4\cdot1+6\cdot4+8\cdot7}{4+2+1+1+4+7}\\\\\\\overline x =\dfrac{-12-4+3+4+24+56}{19}\\\\\\\overline x =\dfrac{71}{19}=3.7368421....\approx3.74$

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Help me quickly please

liberstina [14]

$P_{freshman}=\frac{freshman}{total}=\frac{7}{30}=0.2333\approx23\%$

7 0

3 years ago

Simplify the expression c^-6 * d^4/ c^-16 * d^14

Naily [24]

Answer:

$\dfrac{c^{10}}{d^{10}}$

Step-by-step explanation:

Although the way you wrote problem, this is not what it looks like, I think this is what you meant.

$\dfrac{c^{-6} \times d^{4}}{c^{-16} \times d^{14}} =$

$= \dfrac{c^{-6}}{c^{-16}} \times \dfrac{d^{4}}{d^{14}}$

$= c^{-6 - (-16)} \times d^{4 - 14}$

$= c^{-6 + 16} \times d^{-10}$

$= c^{10} \times d^{-10}$

$= \dfrac{c^{10}}{d^{10}}$

7 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

TRUE OR FALSE: <br> An isosceles trapezoid is a trapezoid with congruent legs.

dem82 [27]

Answer:

True

Step-by-step explanation:

If a quadrilateral (with one set of parallel sides) is an isosceles trapezoid, its legs are congruent.

4 0

3 years ago

Read 2 more answers

a business shows a profit of $110 for selling 20 leather wallets how much profit would the business make if they sold 35 wallets

mafiozo [28]

Answer:

They would get 192.5

Step-by-step explanation:

First we need to find how much it cost per 1 so you will need to dovide 110 by 20 which equal 5.5

Then you will multiply 5.5*35 and get 192.5

8 0

3 years ago