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sdas [7]
2 years ago
5

In 1895, a brick arch railway bridge was built on North Avenue in Baltimore, Maryland. The arch is described by the equation h =

9 −1/50x², where h is the height in yards and x is the distance in yards from the center of the bridge.
a. Graph this equation on your calculator and describe, to the nearest yard, where the
bridge touches the ground.
Mathematics
1 answer:
DerKrebs [107]2 years ago
7 0

Answer:

21.213

or

(21.213,0)

Step-by-step explanation:

y-intercept = 9

slope= -1/50x^2

I graphed it and it was a parabola and it hit the ground at this coordinate (21.213,0)

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The Chang's neighbor owns a ready-mix company. The company receives an order for 12.5 cubic yards of concrete. This mixture of c
ella [17]
The quantities in the proportion you give add to 16 pounds, of which 2 pounds are cement. That is, cement constitutes 1/8 of the total weight.

For each yard of concrete, 4000 lb/8 = 500 lb of cement are used.

For 12.5 yards of concrete, 12.5 * 500 = 6,250 pounds of cement will be used.

_____
"yard" in this case is shorthand for "cubic yard".
3 0
2 years ago
Find a single expression that represents the area of the outer ring of the circle if the area of the whole circle is represented
sp2606 [1]

Answer:  The answer is A_o=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


Step-by-step explanation: Given that the area of the whole circle is represented by the expression

A_c=25x^2-12x-9.

We are to find the area of the outer ring of the circle, i.e., to find the circumference of the circle.

Now, if 'r' represents the radius of the circle, then we have

A_c=\pi r^2\\\\\Rightarrow \dfrac{22}{7}r^2=25x^2-12x-9\\\\ \Rightarrow r^2=\dfrac{7}{22}(25x^2-12x-9)\\\\\Rightarrow r=\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.

Thus, the area of the outer ring is

A_o=2\pi r=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


5 0
2 years ago
Suppose a marketing company computed a 94% confidence interval for the true proportion of customers who click on ads on their sm
Ksivusya [100]

Answer:

d. There is a 98% chance that the true proportion of customers who click on ads on their smartphones is between 0.56 and 0.62.

Step-by-step explanation:

Confidence interval:

x% confidence

Of a sample

Between a and b.

Interpretation: We are x% sure(or there is a x% probability/chance) that the population mean is between a and b.

In this question:

I suppose(due to the options) there was a small typing mistake, and we have a 98% confidence interval between 0.56 and 0.62.

Interpreation: We are 98% sure, or there is a 98% chance, that the true population proportion of customers who click on ads on their smartphones is between 0.56 and 0.62. Option d.

3 0
2 years ago
The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat
Greeley [361]

Answer:

1) L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

2) T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

3) T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

Part 2

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

Part 3

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

Replacing we got:

T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

8 0
2 years ago
PLEASE HELP ME!
dezoksy [38]
Thicknesses at different point are: <span>41, 38, 36, 29, 34, 44, 46, 43, 35, 40


In increasing order: 29, 34, 35, 36, 38, 40, 41, 43, 44, 46

Median = (38+40)/2 = 39m</span>

Median thickness is 39m
5 0
2 years ago
Read 2 more answers
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