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mylen [45]
2 years ago
15

Henry is designing a model for a dome shaped glasshouse. He creates the model and records the horizontal distance from the edge

of the greenhouse of the dome over its vertical height from the base of the model. The table shows the horizontal distance from the edge of the greenhosue, in inches, x, over its verital height from the base, in inches, f(x). Use data in table to create standard form of the function that models this situation.

Mathematics
2 answers:
dolphi86 [110]2 years ago
8 0

Answer:

Correct choice is B

Step-by-step explanation:

All given options represent quadratic function. Let the equation of this quadratic function be

f(x)=ax^2+bx+c.

Then

1. f(0)=2=a\cdot 0^2+b\cdot 0+c\Rightarrow c=2;

2. f(1)=7.5=a\cdot 1^2+b\cdot 1+c\Rightarrow 7.5=a+b+2;

3. f(2)=12=a\cdot 2^2+b\cdot 2+c\Rightarrow 12=4a+2b+2.

Solve the system of two equations:

\left\{\begin{array}{l}a+b+2=7.5\\4a+2b+2=12\end{array}\right.\Rightarrow\left\{\begin{array}{l}a=5.5-b\\4(5.5-b)+2b=10\end{array}\right.

Then

22-4b+2b=10,\\ \\-2b=-12,\\ \\b=6,\\ \\a=5.5-6=-0.5.

Thus, the equation of the function is

f(x)=-\dfrac{1}{2}x^2+6x+2.

Note that

f(3)=-\dfrac{1}{2}\cdot 3^2+6\cdot 3+2=-4.5+20=15.5;

f(4)=-\dfrac{1}{2}\cdot 4^2+6\cdot 4+2=18.

lara31 [8.8K]2 years ago
5 0

Answer:

Step-by-step explanation:

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The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar
Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

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Answer:

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