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il63 [147K]
3 years ago
8

Ryan rented a truck for one day. There was a base fee of $15.99, and there was an additional charge of 72 cents for each mile dr

iven. Ryan had to pay $175.11 when he returned the truck. For how many miles did he drive the truck?
Mathematics
1 answer:
Alexandra [31]3 years ago
6 0

Answer: he drove 221 miles

Step-by-step explanation:

You subtract the initial fee from the total cost (175.11-15.99=159.12) the you divide that by the cost of every mile and it gives you the amount of miles.

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(PHOTO) Please help need this done by tomorrow thank you.​
Andre45 [30]

Answer:

x = 8

Step-by-step explanation:

These are corresponding angles, and they are equal to one another.

7x+3=3x+35\\4x+3=35\\4x=32\\x=8

8 0
2 years ago
A study of five hundred adults found that the number of hours they spend on social networking sites each week is normally distri
Nata [24]
Margin of error formula:
ME = z * (Sigma) / √ n
z = 1.96 for a 95 % confidence interval;
Sigma = 3 ( the population standard deviation )
n = 500
ME = 1.96 * 3 / √500 =
= 1.96 * 3 / 22.36 =
= 5.88 / 22.36 = 0.263
Answer:
The margin error for a 95% confidence interval is 0.263.
6 0
3 years ago
What are the solutions to the expression 4x2+11x−3
lina2011 [118]

Step-by-step explanation:

4x² + 11x - 3 = 0

(4x-1)(x+3) = 0

x = 1/4, -3

6 0
2 years ago
Read 2 more answers
Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground
Scorpion4ik [409]

Answer:

125.4\ \text{m/s}

Step-by-step explanation:

u = Initial velocity of baseball

\theta = Angle of hit = 30^{\circ}

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

y_0 = Height of hit = 2.5 ft

a_y = g = Acceleration due to gravity = 32.2\ \text{ft/s}^2

t = Time taken

Displacement in x direction

x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}

Displacement in y direction

y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}

The minimum initial velocity needed for the ball to clear the fence is 125.4\ \text{m/s}

5 0
3 years ago
Substitute y=-6x-3 y=-x+2
MrRissso [65]
-6x-3= -x+2
-5x=5
x=-1
4 0
3 years ago
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