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marusya05 [52]
3 years ago
5

Write an equation in slope-intercept form for the line with slope -2 and y-intercept 5.

Mathematics
1 answer:
Nana76 [90]3 years ago
5 0

Answer:

y = -2x + 5

Step-by-step explanation:

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6+3.245<br><br> Find the sum
Effectus [21]

Answer:

9.245

Step-by-step explanation:

6 + 3.245 = 9.245

4 0
3 years ago
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Find all solutions of the equation. (enter your answers as a comma-separated list.) x2 + 5x + 9 = 0
erik [133]
X,=-(5±✓25-36) ,/ 2

x={-5±✓-11)/2

x= -5/2 ± i✓11/2
4 0
3 years ago
The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05
Tresset [83]

Answer:

(a) Probability that there are no surface flaws in an auto's interior is 0.6065 .

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws is 0.00673 .

Step-by-step explanation:

We are given that the number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05 flaws per square foot of plastic panel.

Let X = Distribution of number of surface flaws in plastic panels

So, X ~ Poisson(\lambda)

The mean of Poisson distribution is given by, E(X) = \lambda = 0.05

which means, X ~ Poisson(0.05)

The probability distribution function of a Poisson random variable is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}; for  x=0,1,2,3...

Now, we know that \lambda for per square foot of plastic panel is 0.05 and we are given that an automobile interior contains 10 square feet of plastic panel.

Therefore,  for 10 square foot of plastic panel is = 10 * 0.05 = 0.5

(a) Probability that there are no surface flaws in an auto's interior =P(X=0)

    P(X = 0) = \frac{e^{-0.5}*0.5^{0}}{0!} = e^{-0.5} = 0.6065

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws = P(X = 0)^{10}

    So, P(X = 0)^{10} = 0.6065^{10} = 0.00673

8 0
3 years ago
How to find the derivative of e^3㏑(x²)
vovikov84 [41]
f(x)=e^{3\ln x^2}\\\\f'(x)=e^{3\ln x^2}\cdot\dfrac{3}{x^2}\cdot2x=e^{3\ln x^2}\cdot\dfrac{6}{x}=\dfrac{6e^{3\ln x^2}}{x}\\\\\\but\ e^{3\ln x^2}=e^{\ln (x^2)^3}=e^{\ln x^6}=x^6\ where\ x\neq0\\\\therefore\ f'(x)=6x^5
7 0
3 years ago
Determine the mean, median, and mode of the data set.
weqwewe [10]

Answer:

mean=21.7

median=23

mode=12

Step-by-step explanation:

Mean=37+12+32+23+12+40+11+12+10+23+27/11

=21.7(3 s.f)

Median: 10,11,12,12,12,23,23,27,32,37,40

Median=23

Mode=12

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2 years ago
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