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coldgirl [10]
3 years ago
5

Negative 4.2x+31.5=negative 65.1

Mathematics
2 answers:
Rasek [7]3 years ago
7 0
It is a linear simple equation for one unknown, lets solve it!
-4.2x + 31.5 = -65.1
we can easily get rid of decimals if we multiply the whole equation times 10:
<span>-42x + 315 = -651
</span>-42x = -651 - 315 = -966
x = 966/42
x = 23
that is the solution
Deffense [45]3 years ago
6 0

Answer:

its option A in edg 2020

Step-by-step explanation:

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I have this math question that I need an answer to. it's 3/4 - 2/5
Marysya12 [62]

Answer:

7/20

3/4-2/5=7/20

4 0
3 years ago
What is the average rate of change for the quadratic function from x=-1 to x=4?
Mazyrski [523]

Answer:

1

Step-by-step explanation:

f(x) = x^2 - 2x - 4

when x = -1, f(-1) = (-1)^2 - 2(-1) - 4 = 1 + 2 - 4 = -1

when x = 4, f(4) = (4)^2 - 2(4) - 4 = 16 - 8 - 4 = 4

The average rate of change from x = -1 to x = 4:

  f(4) - f(-1)          4 - (-1)              5

---------------- = -----------------=   --------  = 1

 x4 - x1               4 - (-1)              5

Answer

The average rate of change = 1

7 0
3 years ago
7.07x10exponet2+3.51x10exponet3
Lynna [10]

Answer:

4217

Step-by-step explanation:

7.07(10^2)+3.51(10^3)

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707+3.51(10^3)

707+(3.51)(1000)

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4217

4 0
3 years ago
Which congruence theorems or postulate could be given as reason why CDE ~= OPQ
Alla [95]

Answer:

are congruent by ASA property

Step-by-step explanation:

AND TELL ME YOUR CLASS

5 0
3 years ago
Read 2 more answers
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.

B)

- 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
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