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dimaraw [331]
3 years ago
9

In a statistics course, a linear regression equation was computed to predict the final exam score from the score on the first te

st. The equation was y = 10 + .9x where y is the final exam score and x is the score on the first test. Carla scored 95 on the first test. On the final exam, Carla scored 98. What is the value of her residual?
Mathematics
1 answer:
eduard3 years ago
4 0

Answer:

2.5

Step-by-step explanation:

The predicted regression equation is

y^=10+0.9x

where y= final exam score and x=first test score.

Also, we are given that Carla's final exam score y is 98 and Carla's first test score x is 95. We have to find residual for Carla.

Residual=Observed -predicted=y-y^

y^=10+0.9x

We know that Carla's first test score=x=95.

So,

y^=10+0.9(95)

y^=10+85.5

y^=95.5

Residual for Carla=y-y^=?

We know that Carla's final exam score=y=98.

So,

Residual for Carla=98-95.5=2.5

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4 years ago
How many grams of a 15 percent alcohol solution should be mixed with 40 grams of a 30 percent solution to obtain a
Akimi4 [234]

The number of  grams is 20 grams

Let x represent how  many grams of a 15 percent alcohol solution

15%x+30%.40/x+40

=25%

0.15x+0.4.30=0.25 (x+40)

0.15x+12=0.25x+10

0.15x-0.25x=10+12

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Inconclusion The number of  grams is  20 grams

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4 0
3 years ago
Ted and Jude are each saving money each month. After x months, the amount of money, in dollars, that Ted has saved is represente
Nadusha1986 [10]

Answer:

True Statements are -

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

Step-by-step explanation:

Given - Ted and Jude are each saving money each month. After x months, the amount of money, in dollars, that Ted has saved is represented by the

equation T = 40x, and the amount of money that Jude has saved is represented by the equation J = 60x.

To find - Choose all of the statements that are true.

A. Each month, Jude saves two-thirds as much money as Ted saves.

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

C. The amount of money Jude saves each month is $20 more than the amount Ted saves each month.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

Proof -

Given that,

After x months,

Ted saved the amount of money, T(x) = 40x

Jude saved the amount of money, J(x) = 60x

Now,

In 1 month,

Ted saved money, T(1) = 40(1) = 40

Jude saved money, J(1) = 60(1) = 60

So,

In 1st month, Jude saved money 20 more than Ted saved.

Now,

In 2 month,

Ted saved money, T(2) = 40(2) = 80

Jude saved money, J(2) = 60(2) = 120

So,

In 2nd month, Jude saved money 40 more than Ted saved.

Now,

In 3 month,

Ted saved money, T(3) = 40(3) = 120

Jude saved money, J(3) = 60(3) = 180

So,

In 3rd month, Jude saved money 60 more than Ted saved.

So,

Option C is incorrect

Because

In 1st month, Jude saves  is $20 more than the amount Ted saves

In 2nd month, Jude saves  is $40 more than the amount Ted saves

Now,

In 1st month,

Ted saves = 40

and

\frac{2}{3}(40) = 26.67

So, Jude will save = 40 + 26.67 = 66.67

But Jude saves 60

So,

Option A is incorrect.

i.e. A. Each month, Jude saves two-thirds as much money as Ted saves.

Now,

We can see that,

In 3 months, Ted will have saved the money = 120

In 2 months, Jude will have saved the money = 120

So,

Option B is correct.

i.e. B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

Also,

We can see that

In 1st month, Jude saved money 20 more than Ted saved.

In 2nd month, Jude saved money 40 more than Ted saved.

In 3rd month, Jude saved money 60 more than Ted saved.

So,

Option D is correct.

i.e. D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

∴ we get

True Statements are -

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

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3 years ago
A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.
Usimov [2.4K]

Answer:

a) \frac{ds}{dt}= v(t) = 3t^2 -18t +15

b) v(t=3) = 3(3)^2 -18(3) +15=-12

c) t =1s, t=5s

d)  [0,1) \cup (5,\infty)

e) D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

f) a(t) = \frac{dv}{dt}= 6t -18

g) The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

Step-by-step explanation:

For this case we have the following function given:

f(t) = s = t^3 -9t^2 +15 t

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

\frac{ds}{dt}= v(t) = 3t^2 -18t +15

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

v(t=3) = 3(3)^2 -18(3) +15=-12

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

3t^2 -18 t +15 =0

We can divide both sides of the equation by 3 and we got:

t^2 -6t +5=0

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

(t-5)*(t-1) =0

And for this case we got t =1s, t=5s

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

t^2 -6t +5 >0

(t-5) *(t-1)>0

So then the intervals positive are [0,1) \cup (5,\infty)

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6

And if we replace we got:

D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

a(t) = \frac{dv}{dt}= 6t -18

Part g and h

The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

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3 years ago
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