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qaws [65]
2 years ago
6

The fuel efficiency for a 2007 passenger car was 31.2 mi/gal. For the same model of car, the fuel efficiency increased to 35.6 m

i/gal in 2012. The gas tank for this car holds 16 gallons of gas. Write and graph a linear function that models the distance that each car can travel for a given amount of gas (up to one tankful). Write the domain and range of each function using interval notation. Write and simplify a function f(g) that represents the difference in the distance that the 2012 car can travel and the distance that the 2007 car can travel on the same amount of gas. Interpret this function using the graphs of the functions from part a. Also, find and interpret f(16). Write the domain and range of the difference function using set notation.

Mathematics
1 answer:
Ierofanga [76]2 years ago
3 0

We have been given that fuel efficiency for a 2007 passenger car was 31.2 mi/gal and the same model of car, the fuel efficiency increased to 35.6 mi/gal in 2012. Also, the gas tank for this car holds 16 gallons of gas.

We need to write a function and graph a linear function that models the distance that each car can travel for a given amount of gas up to one tankful.

Let represent the functions as d_{1}(x) and d_{2}(x)  where d_{1} and d_{2} represent the distances traveled by car in years 2007 and 2012 and x represents the number of gallons. Therefore, we can express the required functions as:

d_{1}(x)=31.2x \text{ and } d_{2}(x) = 35.6x

Domain of both these functions are [0,16] and ranges are [0,499.2] and [0,569.6] respectively for years 2007 and 2012.

The difference function will be:

f(x)=35.6x-31.2x\\
f(x)=4.2x

f(16)=4.2*16=67.2

Domain of this function is [0,16] and range is [0,67.2].

The graphs are shown below.


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aleksley [76]

Answer:

a. -1.60377

b. 0.25451

c. 0.344

d. Option b) 78th

Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

Option c) .253 is.correct

c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

For 39 Skittles

z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

For 42 Skittles

z = (42 - 38.4)/2.12

= 1.69811

Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

0.95526 - 0.61142

0.34384

= 0.344

Option c is.correct

d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (40 - 38.4)/2.12

= 0.75472

P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

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When graphing the inequality y ≤ 2x − 4, the boundary line needs to be graphed first. Which graph correctly shows the boundary l
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In a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979,
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Answer:

we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years

Step-by-step explanation:

Given that in a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979, two groups of childless wives aged 25 to 29 were selected at random, and each was asked if she eventually planned to have a child. One group was selected from among wives married less than two years and the other from among wives married five years.

Let X be the group married less than 2 years and Y less than 5 years

                         X        Y     Total

Sample size   300   300    600

Favouring       240   288    528

p                      0.8     0.96  0.88

H_0: p_x=p_y\\H_a: p_x>p_y

p difference = -0.16

Std error for difference = \sqrt{0.88*0.12/600} =0.01327

Test statistic = p difference/std error=-6.03

p value <0.000001

Since p is less than alpha 0.05 we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years

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2 years ago
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Genrish500 [490]

Answer:

\frac{1}{10}

Step-by-step explanation:

Let t represent the cost of the torch and b represent the cost of the battery.

The torch and a battery cost £2.50 altogether.

\Rightarrow t+b=2.50

The torch costs £2.00 more than the battery.

\Rightarrow t=b+2.00

We substitute the second equation into the first equation to get;

\Rightarrow b+2.00+b=2.50

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£2.50

\Rightarrow 2b=0.50

\Rightarrow b=0.25

The price of the battery is £0.25

We express this as a fraction of the total cost which is £2.50 to get;

\frac{0.25}{2.5}=\frac{1}{10}

8 0
2 years ago
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