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In-s [12.5K]
2 years ago
10

A new car is purchased for $35, 000 and over time its value depreciates by

Mathematics
1 answer:
jekas [21]2 years ago
7 0
The value after 14 year is 8750 dollar

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How much total wrapping paper does it take to wrap two gifts: one with the dimensions of 6 cm by 10 cm by 9.5 cm and the other w
zhuklara [117]

Answer:

gift 1:

424 cm^2

570 cm^3

gift 2:

1420.5 cm^2

3622.5 cm^3

Step-by-step explanation:

We have to calculate the surface area of these gifts in order to know the amount of wrapping paper.

The formula for the surface area is:

A = 2*ab + 2*ac + 2*bc

Gift 1:

A = 2*6*10 + 2*6*9.5 + 2*10*9.5

A = 424 cm^2

That means that gift 1 needs 424 cm^2 of wrapping paper

Gift 2:

A = 2*14*17.25 + 2*14*15 + 2*17.25*15

A = 1420.5 cm^2

That means that gift 2 needs 1420.5 cm^2 of wrapping paper

Now, to calculate the cubic centimeters of packaging peanuts, we calculate the volume, the volume has as formula:

V = a*b*c  

Gift 1:

V = 6*10*9.5  

V = 570 cm^3

This means that gift 1 needs 570 cm^3 of packaging peanuts

Gift 2:

V = 14*17.25*15  

V = 3622.5 cm^3

That means that gift 2 needs 3622.5 cm^3 of packaging peanuts

5 0
2 years ago
What is the slope of the line that passes through the points (10,5) and (15,20)
Ganezh [65]

Answer:

slope=3

Step-by-step explanation:

6 0
2 years ago
(05.02 LC)
Arte-miy333 [17]

<em>Your Answer: </em>A.) (2 , 5) <u>I also provided the steps.</u>

Hope this helps y'all :)


7 0
3 years ago
Read 2 more answers
This problem uses the teengamb data set in the faraway package. Fit a model with gamble as the response and the other variables
hichkok12 [17]

Answer:

A. 95% confidence interval of gamble amount is (18.78277, 37.70227)

B. The 95% confidence interval of gamble amount is (42.23237, 100.3835)

C. 95% confidence interval of sqrt(gamble) is (3.180676, 4.918371)

D. The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

Step-by-step explanation:

to)

We will see a code with which it can be predicted that an average man with income and verbal score maintains an appropriate 95% CI.

attach (teengamb)

model = lm (bet ~ sex + status + income + verbal)

newdata = data.frame (sex = 0, state = mean (state), income = mean (income), verbal = mean (verbal))

predict (model, new data, interval = "predict")

lwr upr setting

28.24252 -18.51536 75.00039

we can deduce that an average man, with income and verbal score can play 28.24252 times

using the following formula you can obtain the confidence interval for the bet amount of 95%

predict (model, new data, range = "confidence")

lwr upr setting

28.24252 18.78277 37.70227

as a result, the confidence interval of 95% of the bet amount is (18.78277, 37.70227)

b)

Run the following command to predict a man with maximum values ​​for status, income, and verbal score.

newdata1 = data.frame (sex = 0, state = max (state), income = max (income), verbal = max (verbal))

predict (model, new data1, interval = "confidence")

lwr upr setting

71.30794 42.23237 100.3835

we can deduce that a man with the maximum state, income and verbal punctuation is going to bet 71.30794

The 95% confidence interval of the bet amount is (42.23237, 100.3835)

it is observed that the confidence interval is wider for a man in maximum state than for an average man, it is an expected data because the bet value will be higher than the person with maximum state that the average what you carried s that simultaneously The, the standard error and the width of the confidence interval is wider for maximum data values.

(C)

Run the following code for the new model and predict the answer.

model1 = lm (sqrt (bet) ~ sex + status + income + verbal)

we replace:

predict (model1, new data, range = "confidence")

lwr upr setting

4,049523 3,180676 4.918371

The predicted sqrt (bet) is 4.049523. which is equal to the bet amount is 16.39864.

The 95% confidence interval of sqrt (wager) is (3.180676, 4.918371)

(d)

We will see the code to predict women with status = 20, income = 1, verbal = 10.

newdata2 = data.frame (sex = 1, state = 20, income = 1, verbal = 10)

predict (model1, new data2, interval = "confidence")

lwr upr setting

-2.08648 -4.445937 0.272978

The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

4 0
3 years ago
6. Which of the following is the solution set to the equation x² + 8x -15 = 5x +13?
Verizon [17]

Answer:

The answer is A.......

5 0
3 years ago
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