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2 years ago

Find the vector equation for the line of intersection of the planes 5x+3y−2z=−45x+3y−2z=−4 and 5x+5z=0

Mathematics

1 answer:

olga_2 [115]2 years ago

8 0

$5x+5z=0\implies x+z=0\implies z=-x$

$5x+3y-2z=7x+3y-2x-2z=7x+3y-2=-4\implies7x+3y=-2$

Letting $x=t$ and $z=-t$ , we have

$7t+3y=-2\implies y=-\dfrac73t-\dfrac23$

so the intersection is given by

$\mathbf r(t)=\left(t,-\dfrac73t-\dfrac23,-t\right)$

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How do I find the area of a circle when the circumference is 58 pie

aksik [14]

C=2 $\pi$ r
A= $\pi$ r²

C=2 $\pi$ r=58 $\pi$
2r=58
r=29

A= $\pi$ r²
A=29² $\pi$
A=841 $\pi$

8 0

2 years ago

Read 2 more answers

Question 5

viva [34]

Marlee will have 21.05 inches of wire left, here is why;
115-25.75=89.25
89.25+30=119.25
119.25-38=81.25
81.25-60.2=21.05
She has 21.05 inches of wire left.

4 0

2 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

2 years ago

The length of a banquet table is 4 feet longer than 6 times its width. If the perimeter is 78 feet, what are the dimensions of t

harina [27]

Hello,
2(l+w)=78
Divided 2 from both side
2(l+w)/2=78/2
l+w=39 feet

l=6w+4
Replace l with 6w+4

6w+4+w=39
7w+4=39
Subtract 4 from both side
7w=35
Divided 7 from both side
w=5 feet
l=6w+4
l=6(5)+4
l=30+4
l=34 feet. As a result, the length of the rectangle is 34 feet and the width of rectangle is 5 feet. Hope it help!

7 0

3 years ago

After robbing a bank in dodge city, a robber gallops off at 14 mi/h. 20 minutes later, the marshall leaves to pursue the robber

Morgarella [4.7K]

For this case we have to write the equation of the position of each one.

We have then:

For the robber:

$r = 14t$