These three vectors are linearly <u>dependent</u> (details are in the attachment).
Answer:
look at the explanation
<em><u>explanation</u></em>
example
6 × 3/4
do 6 ×3
6×4
u will get 18/24 and then simplify
Answer:
Part 1) The product of 3 and 2y-4x=8 is 
Part 2) The original equation in slope intercept form is 
Part 3) The new equation in slope intercept form is 
Step-by-step explanation:
we have

step 1
Find out the product of 3 and 2y-4x=8
Multiply both sides by 3

Apply distributive property both sides

step 2
Find out the original equation in slope intercept form
The equation of the line in slope intercept form is

we have

Solve for y
That means ----> Isolate the variable y
Adds 4x both sides

Divide by 2 both sides

Simplify

step 3
Find out the new equation in slope intercept form
The equation of the line in slope intercept form is

we have

Solve for y
That means ----> Isolate the variable y
Adds 12x both sides

Divide by 6 both sides

Simplify

Answer:
1/2×10×6=30
Step-by-step explanation:
H=6
L=10
therefore;
1/2×6×10=30
Answer:
a) 0.057
b) 0.5234
c) 0.4766
Step-by-step explanation:
a)
To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula
where
N = size of the sample.
So,
As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.
We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.
So the p-value is
b)
Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.
We can compute the probability of such an error following the next steps:
<u>Step 1
</u>
Compute
So <em>we would make a Type II error if our sample mean is less than 185.3721</em>.
<u>Step 2</u>
Compute the probability that your sample mean is less than 185.3711
So, <em>the probability of making a Type II error is 0.5234 = 52.34%
</em>
c)
<em>The power of a hypothesis test is 1 minus the probability of a Type II error</em>. So, the power of the test is
1 - 0.5234 = 0.4766