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3 years ago

Can you answer question 45 and 46? (Show work)

Mathematics

1 answer:

muminat3 years ago

4 0

45. These word problems often seem to be two equations in two unknowns, here h for hits and m for misses.

"After 30 tries" translates into

$h + m = 30$

+5 for a hit, -10 for a miss, John lost 90 translates to

$5h + -10 m = -90$

We have

$m = 30 - h$

$5h - 10(30 -h) = -90$

$5h - 300 + 10h = -90$

$15h = 210$

$h = 210/15 = 14$

Answer: 14

Check: 5(14)-10(16)=-90, good

46. We have to make the odd assumption the grocer sells for his cost, but let's not beat up the question too hard.

Again, two unknowns, b kilograms of Brazil nuts and c kilograms of cashews.

... sold 12 kg of the mixture ...

$b + c = 12$

... cashews $8/kg and Brazil nuts $10/kg and the results is $8.50/kg ...

This one's a bit more involved to translate because we have to recognize the total price is the 8.50 per kg times 12 kg.

$8c + 10b = 12(8.50) = 102.00$

The solving is pretty much just like the last one; we eliminate the unknown we're not asked for.

$b = 12 - c$

$8c + 10(12 - c) = 102$

$-2c = 102 - 120 = -18$

$c = 9$

Answer: 9

Check: (9(8)+10(3))/12=102/12 = 8.50 good

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a triangular pyramid has a volume 150 cubic centimeters. which of the following represents the volume of a triangular prism with

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3 years ago

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2 years ago

1. Answer the following questions. (a) Check whether or not each of f1(x), f2(x) is a legitimate probability density function f1

kobusy [5.1K]

Answer:

f1 ( x ) valid pdf . f2 ( x ) is invalid pdf

k = 1 / 18 , i ) 0.6133 , ii ) 0.84792

Step-by-step explanation:

Solution:-

A) The two pdfs ( f1 ( x ) and f2 ( x ) ) are given as follows:

$f_1(x) = \left \{ {{0.5(3x-x^3) } .. 0 < x < 2 \atop {0} } \right. \\\\f_2(x) = \left \{ {{0.3(3x-x^2) } .. 0 < x < 2 \atop {0} } \right. \\$

- To check the legitimacy of a continuous probability density function the area under the curve over the domain must be equal to 1. In other words the following:

$\int\limits^a_b {f_1( x )} \, dx = 1\\\\ \int\limits^a_b {f_2( x )} \, dx = 1\\$

- We will perform integration of each given pdf as follows:

$\int\limits^a_b {f_1(x)} \, dx = \int\limits^2_0 {0.5(3x - x^3 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - 0.125x^4 ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - 0.125*(16) ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 2 ] = 1\\$

$\int\limits^a_b {f_2(x)} \, dx = \int\limits^2_0 {0.5(3x - x^2 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - \frac{x^3}{6} ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - \frac{(8)}{6} ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 1.3333 ] = 1.67 \neq 1 \\$

Answer: f1 ( x ) is a valid pdf; however, f2 ( x ) is not a valid pdf.

- A random variable ( X ) denotes the resistance of a randomly chosen resistor, and the pdf is given as follows:

$f ( x ) = kx$ if 8 ≤ x ≤ 10

0 otherwise.

- To determine the value of ( k ) we will impose the condition of validity of a probability function as follows:

$\int\limits^a_b {f(x)} \, dx = 1\\$

- Evaluate the integral as follows:

$\int\limits^1_8 {kx} \, dx = 1\\\\\frac{kx^2}{2} ]\limits^1^0_8 = 1\\\\k* [ 10^2 - 8^2 ] = 2\\\\k = \frac{2}{36} = \frac{1}{18}$ ... Answer

- To determine the CDF of the given probability distribution we will integrate the pdf from the initial point ( 8 ) to a respective value ( x ) as follows:

$cdf = F ( x ) = \int\limits^x_8 {f(x)} \, dx\\\\F ( x ) = \int\limits^x_8 {\frac{x}{18} } \, dx\\\\ F ( x ) = [ \frac{x^2}{36} ] \limits^x_8\\\\F ( x ) = \frac{x^2 - 64}{36}$

To determine the probability p ( 8.6 ≤ x ≤ 9.8 ) we will utilize the cdf as follows:

p ( 8.6 ≤ x ≤ 9.8 ) = F ( 9.8 ) - F ( 8.6 )

p ( 8.6 ≤ x ≤ 9.8 ) = $\frac{(9.8)^2 - 64}{36} - \frac{(8.6)^2 - 64}{36} = 0.61333$

ii) To determine the conditional probability we will utilize the basic formula as follows:

p ( x ≤ 9.8 | x ≥ 8.6 ) = p ( 8.6 ≤ x ≤ 9.8 ) / p ( x ≥ 8.6 )

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 1 - p ( x ≤ 8.6 ) ]

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 1 - 0.27666 ]

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 0.72333 ]

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.84792 ... answer

3 0

3 years ago