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3 years ago

15

An auto-parts store offers a fuel additive that claims to increase a vehicle’s gas mileage. The additive is poured into a vehicl

e’s gasoline tank after the tank is filled. To measure the claim, 30 taxi drivers receive a bottle of the additive to use the next time they fill their gas tank. The drivers then compare their gas mileage with the additive to their gas mileage before the additive was used.
What is the response variable in this scenario?

the age of the car
the 30 cars used in the study
the use of the additive
the gas mileage after the additive is used

2 answers:

AleksAgata [21]3 years ago

3 0

Answer:

The answer is D. the gas mileage after the additive is used

Step-by-step explanation:

I got 100% on edg!

tekilochka [14]3 years ago

3 0

Answer:

d

Step-by-step explanation:

edge2o2o

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Solve |3k-2|=2|k+12|

Stella [2.4K]

Answer:

$k = 26$ or $k = -\dfrac{22}{5}$

Step-by-step explanation:

|3k - 2| = 2|k + 12|

$\dfrac{|3k - 2|}{|k + 12|} = 2$

$|\dfrac{3k - 2}{k + 12}| = 2$

$3k - 2 = 2(k + 12)$ or $3k - 2 = -2(k + 12)$

$3k - 2 = 2k + 24$ or $3k - 2 = -2k - 24$

$k = 26$ or $5k = -22$

$k = 26$ or $k = -\dfrac{22}{5}$

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3 years ago

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

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3 years ago

CAN SOMEONE HELP ME WITH THIS PLEASE AND THANK YOU

Arturiano [62]

3/4 times 5/8= 15/32

The answer is 15/32

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3 years ago

A train traveled the first 200 miles of its trip at a speed of 50 mph and the next 130 miles at a speed of 65 mph. What was the

12345 [234]

55 miles per hour, to answer questions like these you need to calculate the time (T=D/R) for each individual portion of the trip, add the time together to get total time, and then divide the total distance by the total time in order to get the average velocity in mph.

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3 years ago

Need help asap isnt hard i just dont understand

qwelly [4]

Answer:

The answer is b. 35.5

Step-by-step explanation:

angle bisector always bisects an angle in to half amd in the question you have been told to tell the value of one half of the angle abc so the half of 71 would be 35.5

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3 years ago