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Vitek1552 [10]
1 year ago
7

A parabolic arch sculpture is on top of a city bank. A model of the arch is y = −0.005x2 + 0.3x where x and y are in feet.

Mathematics
1 answer:
o-na [289]1 year ago
7 0

Answer:

A. a. 34.5 feet

b. 60 feet

Step-by-step explanation:

Parabola equation

y=-0.005x^2+0.3x

Differentiating with respect to x we get

\dfrac{dy}{dx}=-0.01x+0.3

Equating with zero

-0.01x+0.3=0\\\Rightarrow -0.01x=-0.3\\\Rightarrow x=\dfrac{0.3}{0.01}\\\Rightarrow x=30

Double derivative of the parabolic equation

\dfrac{d^2y}{dx^2}=-0.01

So, x=30 is maximum.

y=-0.005\times 30^2+0.3\times 30\\\Rightarrow y=4.5

So, the maximum height of the arch will be 4.5 feet.

From the ground the highest point of the arch will be 30+4.5=34.5\ \text{ft}

We are taking the x axis as the width of the bank.

0=-0.005x^2+0.3x\\\Rightarrow 0.005x^2=0.3x\\\Rightarrow x=\dfrac{0.3}{0.005}\\\Rightarrow x=60

So, the width of the bank will be 60 feet.

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PQ=2x+1 and QR=5x-44; Find PR
Dima020 [189]

Answer:

x=15

Step-by-step explanation:

subtract 1 from both sides so you are left with 2x=5x-45

then subtract 5x from both sides and you have -3x=-45

then finally divide -3 from -45 to get x=15

4 0
1 year ago
From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def
jasenka [17]

Answer:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

Sd(X) = \sqrt{0.3075}= 0.5545

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X         0           1           2         3

P(X)    0.92     0.03    0.03     0.02

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

Sd(X) = \sqrt{0.3075}= 0.5545

8 0
2 years ago
Torrey starts a new job with an annual salary of $60,000. For each year she continues to work for the same company, she will rec
GarryVolchara [31]
The third choice will work
6 0
1 year ago
Read 2 more answers
A city that had 40,000 trees started losing them at a rate of 10 percent per year because of urbanization.
goblinko [34]

Answer:

t = 13.1576  years

Step-by-step explanation:

The situation can be modeled as

P = Po.(1-r)^t

Where

t is the years transcurred

Po is the initial amount

r is the rate of change

Po = 40000

r = 10% equivalent to 0.1

Now

A quarter of the original amount

(1/4)*Po = Po.(1-r)^t

(1/4) = (0.9)^t

t = 13.1576

Please, see attached picture

7 0
1 year ago
The probability that a child will learn to swim before age 6 is 0.312. If you take a group of 12 children, what is the probabili
vagabundo [1.1K]

Answer:

The probability of 1 or less children from that group to learn how to swim before 6 years of age is 0.072

Step-by-step explanation:

In this case we need to compute the probability of none of these 12 children learns to swim before 6 years of age. This is given by:

p(0) = (1 - 0.312)^(12) = 0.688^(12) = 0.01124

We now need to calculate the probability that one child learns to swim before 6 years of age.

p(1) = 12*0.312*(1 - 0.312)^(11) = 3.744*(0.688)^(11)

p(1) = 3.744*0.01634

p(1) = 0.0612

The probability of 1 or less children from that group to learn how to swim before 6 years of age is:

p = p(0) + p(1) = 0.01124 + 0.0612 = 0.07244

6 0
2 years ago
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