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V125BC [204]
2 years ago
15

If you bought a stock last year for a price of 103$ and it has risen 13.5% since then, how much is the stock worth now, to the n

earest cent?​
Mathematics
1 answer:
DIA [1.3K]2 years ago
7 0

Answer:

$90.75

Step-by-step explanation:

103*100=(13.5%+100%)*x

10300=113.5x

x=10300/113.5

x=90.74889868

x=$90.75

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Evaluate 8j -k+14 when j = 0.25 and k =1
Amiraneli [1.4K]
Substitute
8(0.25) -1+14 = 15
7 0
3 years ago
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Use the net to find the surface area of the cylinder.
Flauer [41]
Surfacea are=2circles+rectangle

areacircles=2*pi*5^2=2*pi*25=50pi

rectangle, we need circunference of circle=2pir=2pi5=10pi
area=10pi times 7=70pi

add
50pi+70pi=120pi yd^2
 
4 0
3 years ago
To do so, the research has 100 high school students complete the maze with classical music playing. The mean time for these stud
Marizza181 [45]

Answer:

test statistic, z =  −1.74

Step-by-step explanation:

given data

students n = 100

mean time x  = 41.13 seconds

σ  = 5

we consider µ = 42

solution

we get here The test statistic that is express as

test statistic, z = \frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}      ......................1

put here value and we will get

test statistic, z = \frac{41.13-42}{\frac{5 }{\sqrt{100}}}

test statistic, z =  −1.74

8 0
3 years ago
Which simpler problems could be calculated to solve this problem?
Romashka-Z-Leto [24]
The answer would be B.
6 0
3 years ago
Read 2 more answers
A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.
sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}

Where:

y_{o} - Initial height of the ball, measured in feet.

v_{o} - Initial speed of the ball, measured in feet per second.

g - Gravitational constant, equal to -32.174\,\frac{ft}{s^{2}}.

t - Time, measured in seconds.

Given that y_{o} = 0\,ft, v_{o} = 95\,\frac{ft}{s}, g = -32.174\,\frac{ft}{s^{2}} and y = 120\,ft, the following second-order polynomial is found:

120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}

-16.087\cdot t^{2} + 95\cdot t -120 =0

The roots of this polynomial are, respectively:

t_{1} \approx 4.075\,s and t_{2} \approx 1.831\,s.

Both roots solutions are physically reasonable, since t_{1} represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas t_{2} represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

8 0
3 years ago
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