Answer: A is correct i think
Step-by-step explanation: 6a x 5 = 30a
6a x d = 6ad
Answer:
A
Step-by-step explanation:
A national consumer agency selected independent random samples of 45 owners of newer cars (less than five years old) and 40 owners of older cars (more than five years old) to estimate the difference in mean dollar cost of yearly routine maintenance, such as oil changes, tire rotations, filters, and wiper blades. The agency found the mean dollar cost per year for newer cars was $195 with a standard deviation of $46. For older cars, the mean was $286 with a standard deviation of $58. Which of the following represents the 95 percent confidence interval to estimate the difference (newer minus older) in the mean dollar cost of routine maintenance between newer and older cars?
A. (195 - 286) + 1.992 underroot46/45 + 58/40
B. (286 - 195) + 1.992 underroot46^2/45+ 58^2/40
C. (195 - 286)+ 1.992 underrot 140² 158²/45+40
D. (286 - 195) + 1.992 46 1582 45 +40
E. (195-286) +1.992 462 + 58 45 40
Answer:
(195 - 286) ± 1.992 * sqrt((46^2/45) + (58^2/40))
Step-by-step explanation:
Given that:
NEWER CARS:
Sample size = n1 = 45
Standard deviation s1 = 46
Mean = m1 = 195
OLDER CARS:
Sample size = n2 = 40
Standard DEVIATION s2 = 58
Mean = m2 = 286
Confidence interval at 95% ; α = 1 - 0.95 = 0.05 ; 0.05 / 2 = 0.025
Confidence interval is calculated thus : (newer--older)
(m1 - m2) ± Tcritical * standard error
Mean difference = m1 - m2; (195 - 286)
Tcritical = Tn1+n2-2, α/2 = T(45+40)-2 = T83, 0.025 = 1.99 (T value calculator)
Standard error (E) = sqrt((s1²/n1) + (s2²/n2))
E = sqrt((46^2/45) + (58^2/40))
Hence, confidence interval:
(195 - 286) ± 1.992 * sqrt((46^2/45) + (58^2/40))
The 95% confidence interval for estimating the difference in the mean dollar cost of the routine maintenance between newer and older cars is given by:
[tex]CI=(\overline{x_1} - \overline{x_2}) \pm t_{critical} \sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}[/tex]
Given that:The first sample consists of owners of newer cars
First sample's size = [tex]n_1 =45[/tex]First sample's mean = [tex]\overline{x_1}=\$195[/tex]First sample's standard deviation = [tex]s_1 = \$46[/tex]The second sample consists of owner of older cars.
Second sample's size = [tex]n_2 = 40[/tex]Second sample's mean = [tex]\overline{x_2}=\$286[/tex]Second sample's standard deviation = [tex]s_2 = \$58[/tex]To find:95% confidence interval for difference between both samples' means.
Calculations and Explanations:Since the sample sizes are > 30, thus we can use the z table for finding the Confidence interval.
The CI is given as:
[tex]CI=(\overline{x_1} - \overline{x_2}) \pm t_{critical} \sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}[/tex]
For 0.95 probability confidence, we have t at 40+45-3= t at 83 at 0.05/2 is 1.992 (from T tables)
Thus,
[tex]CI=(195-268) \pm 1.992 \sqrt{\dfrac{46^2}{45} + \dfrac{58^2}{40}}[/tex]
Learn more about confidence interval here:
https://brainly.com/question/4249560
A music store marks up the instruments it sells by 30%.
If the price tag on a trumpet says $104, how much did the store pay for it?
A. $ 58.50
B. $ 80
C. No, it did not mark up 30%.
Step-by-step explanation:Since,
A. We have,
Mark up percentage = 30%
Actual price = $ 45,
B. We have,
SP = $ 104,
C. We have,
AP = $ 75,
SP = $100,
Thus, mark up percentage =
= 33.33 %
Hence, store did not mark up the price by 30%.
The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years. a. Find the probability that a cat will live to be older than 14 years. b. Find the probability that a cat will live between 14 and 18 years. c. If a cat lives to be over 18 years, would that be unusual? Why or why not? d. How old would a cat have to be to be older than 90% of other cats?
Answer
a. 0.856
b. 0.78071
c. It is not unusual
d. 13.65 years old
Step-by-step explanation:
The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years.
We solve this question using z score formula:
z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
a. Find the probability that a cat will live to be older than 14 years.
For x > 14 years
z = 14 - 15.7/1.6
z = -1.0625
Probability value from Z-Table:
P(x<14) = 0.144
P(x>14) = 1 - P(x<14) = 0.856
b. Find the probability that a cat will live between 14 and 18 years.
For x = 14 years
z = 14 - 15.7/1.6
z = -1.0625
Probability value from Z-Table:
P(x = 14) = 0.144
For x = 18 years
z = 18 - 15.7/1.6
z= 1.4375
Probability value from Z-Table:
P(x = 18) = 0.92471
The probability that a cat will live between 14 and 18 years is calculated as:
P(x = 18) - P(x = 14)
0.92471 - 0.144
= 0.78071
c. If a cat lives to be over 18 years, would that be unusual? Why or why not?
For x > 18 years
z = 18 - 15.7/1.6
z= 1.4375
Probability value from Z-Table:
P(x<18) = 0.92471
P(x>18) = 1 - P(x<18) = 0.075288
Converting this to percentage:
0.075288 × 100 = 7.5288%
Hence, 7.5288% of the cats live to be over 18 years. Hence, it is not unusual.
d. How old would a cat have to be to be older than 90% of other cats?
From the question above, 10% of the cats would be older than 90% of other cats.
Hence, we find the z score of the 10th percentile
= -1.282
Hence,
-1.282 = x - 15.7/1.6
Cross Multiply
-1.282 × 1.6 = x - 15.7
- 2.0512 = x - 15.7
x = 15.7 -2.0512
x = 13.6488 years old
Approximately = 13.65 years old
Pls help me with this ASAP