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2 years ago

15

A local game store earns a profit of $15.50 for every game they sell. The profit from every game they sell is deposited directly

into their savings account which currently has a balance of $469. If their goal is to increase their balance to $825.50, how many games must the store sell? (3 points)
53

23

84

30

1 answer:

Nikolay [14]2 years ago

3 0

Answer:

B. 23 games.

Step-by-step explanation:

Current balance: $469

Goal balance: $825.50

Profit per game: $15.50

825.50 - 469 = 356.50

They only need $365.50 in order to reach the goal.

356.50 ÷ 15.50 = 23

They need to sell 23 games, each selling for $15.50, in order to reach their goal balance, of $825.50.

Therefore, the answer is B.

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The Solution:

Given:

Required:

Find the standard deviation of the probability distribution.

Step 1:

Find the expected value of the probability distribution.

$E(x)=\mu=\sum_{i\mathop{=}0}^3x_iP_(x_i)$ $\begin{gathered} \mu=(0\times0.25)+(1\times0.05)+(2\times0.15)+(3\times0.55) \\ \\ \mu=0+0.05+0.30+1.65=2.0 \end{gathered}$

Step 2:

Find the standard deviation.

$Standard\text{ Deviation}=\sqrt{\sum_{i\mathop{=}0}^3(x_i-\mu)^2P_(x_i)}$ $=(0-2)^2(0.25)+(1-2)^2(0.05)+(2-2)^2(0.15)+(3-2)^2(0.55)$ $=4(0.25)+1(0.05)+0(0.15)+1(0.55)$ $=1+0.05+0+0.55=1.60$

Thus, the standard deviation is 1.60

Answer:

1.60

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Ray of sun is a line

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2 years ago

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A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt

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Answer:

$2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}$

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be $y$ . The magnitude of $40\text{ m/s}$ will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is $60^{\circ}$ , we have:

$\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}$ (Recall that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ )

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation $v_f^2=v_i^2+2a\Delta y$ to solve this problem, where $v_f/v_i$ is final and initial velocity, respectively, $a$ is acceleration, and $\Delta y$ is distance travelled. The only acceleration is acceleration due to gravity, which is approximately $9.8\:\mathrm{m/s^2}$ . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

What we know:

$v_i=20\sqrt{3}\text{ m/s}$
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A truck can be rented from Company A for $40 a day plus $0.40 per mile. Company B charges $20 a day plus $0.50 per mile to r

torisob [31]

The number of miles in day at which the rental cost for the company A and Company B are the same is 200 miles

The Company A rented the truck for $40 a day plus $0.40 per miles

The expression will be

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Where x is the number of miles

The company B rented the truck for $20 a day plus $0.50 per miles

20+0.50x

To find the number of miles in a day at which the rental costs for Company A and Company B are the same, the linear equation will be

40+0.40x = 20+0.50x

40-20 = 0.50x-0.40x

0.1x = 20

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Hence, the number of miles in day at which the rental cost for the company A and Company B are the same is 200 miles

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