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3 years ago

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Describe the solutions of the following system in parametric vector form,and provide a geometric comparison with the solution se

t .
x1 + 3x2- 5x3 = 4
x1+ 4x2 - 8x3 = 7
-3x1- 7x2 +9x3 =6

1 answer:

aivan3 [116]3 years ago

4 0

Answer:

The equations are linearly independent so there is no parametric vector form

Step-by-step explanation:

I attached the solution.

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nexus9112 [7]

Answer: 104

Step-by-step explanation:

5 0

3 years ago

nolan uses 7 inches of string to make each bracelet. if nolan makes 3 bracelets, how many inches of string will he use?

pantera1 [17]

Answer:

21 inches

Step-by-step explanation:

We can write a ratio to solve

7 inches x inches

------------- = -------------------

1 bracelet 3 bracelets

Using cross products

7*3 = 1x

21 = x

21 inches

6 0

3 years ago

Read 2 more answers

Graph the system of equations.<br><br> {8x+8y=642x−2y=−4

klemol [59]

<span>The graph is attached.

Explanation:
We can use the x- and y-intercepts to graph. The x-intercept of the first equation is 8, and the y-intercept is 8. The x-intercept of the second equation is -2, and the y-intercept is 2.
<span>
x-intercepts are where the data crosses the x-axis. At every one of these points, the y-coordinate will be 0; therefore we can substitute 0 for y and solve to get the value of the x-intercept.

For the first equation, we would have
8x+8(0)=64
8x=64.

Divide both sides by 8:
8x/8 = 64/8
x=8.

For the second equation,
2x-2(0)=-4
2x=-4.

Divide both sides by 2:
2x/2 = -4/2
x=-2.

y-intercepts are where the data crosses the y-axis. At every one of these points, the x-coordinate will be 0; therefore we can substitute 0 for x and solve to get the value of the y-intercept.

For the first equation,
8(0)+8y=64
8y=64.

Divide both sides by 8:
8y/8 = 64/8
y=8.

For the second equation,
2(0)-2y=-4
-2y=-4.

Divide both sides by -2:
-2y/-2 = -4/-2
y=2.

Plot these points for both equations and connect them to draw the line.</span></span>

6 0

3 years ago

6/7 x 56 in multiplied and in its simpliest form

vazorg [7]

So, 6/7*56/1
56*6/7
336/7
the answer is 48

6 0

3 years ago

Read 2 more answers

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, 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

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

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

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

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

3 years ago