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atroni [7]
4 years ago
7

A 4-column table with 5 rows. The first column is labeled age (years) with entries 1, 2, 3, 4, 5. The second column is labeled g

iven value with entries 15, 12, 9, 5, 4. The third column is labeled predicted with entries 14.8, 11.9, 9, 6.1, 3.2. The fourth column is labeled residual with entries 0.2, 0.1, 0, negative 1.1, 0.8.
Which point would be on a residual plot of the data?


(1, 15)
(2, 11.9)
(3, 0)
(5, 21)
Mathematics
2 answers:
andreyandreev [35.5K]4 years ago
6 0

Answer:

It is (3,0)

Step-by-step explanation:

Vesna [10]4 years ago
4 0

Answer:

C

Step-by-step explanation:

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There is a bag filled with 5 blue, 2 red and 3 green marbles,
nikdorinn [45]

Answer:

0.62

Step-by-step explanation:

There are a total of 10 marbles in the bag.

P(blue and not blue) = 5/10 × 5/10 = 25/100

P(red and not red) = 2/10 × 8/10 = 16/100

P(green and not green) = 3/10 × 7/10 = 21/100

P = 25/100 + 16/100 + 21/100

P = 62/100

P = 0.62

Alternatively, you can calculate it this way:

P(both blue) = (5/10)² = 25/100

P(both red) = (2/10)² = 4/100

P(both green) = (3/10)² = 9/100

P = 1 − (25/100 + 4/10 + 9/100)

P = 1 − 38/100

P = 62/100

P = 0.62

5 0
3 years ago
(x+1) - 4<br><br>How do i determine the roots ​
irga5000 [103]
<h2><u>Solution</u>:</h2>

<em>To find the zero,</em>

<em>x</em> + 1 - 4 = 0

<em>x</em> = 4 - 1

<em>x</em> = 3

3 is the <u>root</u> of <em>(x + 1) - 4</em>.

4 0
3 years ago
Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.
const2013 [10]

Answer:

a.

T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) |  =   (-cos(t^2) , -sin(t^2))

b.

T(t) =r'(t)/|r'(t)| =  (2t/ \sqrt{4t^2   + 36} , -6/\sqrt{4t^2   + 36},0)

N(t) =T(t)/|T'(t)| =  (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)

Step-by-step explanation:

Remember that for any curve      r(t)  

The tangent vector is given by

T(t) = \frac{r'(t) }{| r'(t)| }

And the normal vector is given by

N(t) = \frac{T'(t)}{|T'(t)|}

a.

For this case, using the chain rule

r'(t) = (  -10*2tsin(t^2) ,   102t cos(t^2)   )\\

And also remember that

|r'(t)| = \sqrt{(-10*2tsin(t^2))^2  +  ( 10*2t cos(t^2) )^2} \\\\       = \sqrt{400 t^2*(  sin(t^2)^2  +  cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t

Therefore

T(t) = r'(t) / |r'(t) | =  (  -10*2tsin(t^2) ,   10*2t cos(t^2)   )/ 20t\\\\ = (  -10*2tsin(t^2)/ 20t ,   10*2t cos(t^2) / 20t  )\\= ( -sin(t^2), cos(t^2) )

Similarly, using the quotient rule and the chain rule

T'(t) = ( -2t cos(t^2) , -2t sin(t^2))

And also

|T'(t)| = \sqrt{  ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t

Therefore

N(t) = T'(t) / |T'(t) |  =   (-cos(t^2) , -sin(t^2))

Notice that

1.   |N(t)| = |T(t) | = \sqrt{ cos(t^2)^2  + sin(t^2)^2 } = \sqrt{1} =  1

2.   N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0

b.

Simlarly

r'(t) = (2t,-6,0) \\

and

|r'(t)| = \sqrt{(2t)^2   + 6^2} = \sqrt{4t^2   + 36}

Therefore

T(t) =r'(t)/|r'(t)| =  (2t/ \sqrt{4t^2   + 36} , -6/\sqrt{4t^2   + 36},0)

Then

T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)

and also

|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 +   ( (3 t)/(9 + t^2)^{3/2})^2  +  0^2 }\\= 3/(t^2 + 9 )

And since

N(t) =T(t)/|T'(t)| =  (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)

6 0
3 years ago
How much interest would $2,000 earn, with simple interest, in two years at the<br>rate of 4.2%?​
Anna [14]

Answer:

$168

Step-by-step explanation:

The formula we'll use for this is the simple interest formula, or:

I=Prt

Where:

P is the principal amount, $2000.00.

r is the interest rate, 4.2% per year, or in decimal form, 4.2/100=0.042.

t is the time involved, 2....year(s) time periods.

So, t is 2....year time periods.

To find the simple interest, we multiply 2000 × 0.042 × 2 to get that:

The interest is: $168.00

8 0
3 years ago
Kona’s mother realized that her age is the difference of Kona’s age squared and four. If k represents Kona’s age, which expressi
san4es73 [151]
(K²+4)= her moms age :)
7 0
3 years ago
Read 2 more answers
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