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3 years ago

Add the two expressions:

Mathematics

2 answers:

Korvikt [17]3 years ago

4 0

(-6.3x + 4) + (1.5x - 6)

-6.3x + 1.5x = -4.8x
4 + (-6) = -2

-4.8x - 2 is your answer

hope this helps

PolarNik [594]3 years ago

4 0

Answer:

$-4.8x-2$

Step-by-step explanation:

Given : $-6.3x+4\\1.5x-6$

To Solve : Add the given expressions.

Solution :

Expression 1 : $-6.3x+4$

Expression 1 : $1.5x-6$

Add these two expressions :

⇒ $-6.3x+4 +(1.5x-6)$

⇒ $-6.3x+4 +1.5x-6$

⇒ $-4.8x-2$

Thus , The resultant is $-4.8x-2$

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Answer: The true statements are:-

There are about the same number of points above the x-axis as below it.

The points are randomly scattered with no clear pattern.

The number of points is equal to those in the scatter plot.

Explanation:

A residual plot is a graph which shows that the residuals on the vertical axis and the independent variable on the horizontal axis.

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Step-by-step explanation:

You want to know the rate of descent in feet per second represented by a descent of 40 feet in 4 seconds.

The rate is the amount divided by the time:

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2 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

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Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

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Answer:

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