Marti’s car has a sale price of $6,630. How much money does she pay after a $1,090 rebate?

Consider two populations of coins, one of pennies and one of quarters. A random sample of 25 pennies was selected, and the mean

Alexus [3.1K]

Answer:

No, the conditions for normality have not been met because the sample size for the quarters is not large enough and no information is given about the distributions of the populations.

Step-by-step explanation:

"The Kolmogorov-Smirnov test (K-S) and Shapiro-Wilk (S-W) test are designed to test normality by comparing your data to a normal distribution with the same mean and standard deviation of your sample. If the test is NOT significant, then the data are normal, so any value above .05 indicates normality"

5 0

3 years ago

Read 2 more answers

zasha spent $6 on packages of gum. How many more packages of gum that cost $1.20 each can she buy if she has a $20 bill? in a eq

IRISSAK [1]

Answer:

17. (20.4)

Step-by-step explanation:

x times 1.20=20

x is how many packages

5 0

3 years ago

A car travels along a straight stretch of road. It proceeds for 16.2 mi at 57 mi/h, then 24.6 mi at 44 mi/h, and finally 45.1 mi

LekaFEV [45]

Answer:

The average velocity the entire trip was 43.25 mi/h

Step-by-step explanation:

In this case we have to calculate an average based on the miles traveled. First we have to calculate the total of miles traveled, then calculate the portion of the total travel of each and with this calculate the average speed during the trip. First the total miles traveled:

$TotalMiles=16.2+24.6+45.1=85.9$

Now the percentages:

At 57 mi/h were $\frac{16.2}{85.9}=0.19$

At 44 mi/h were $\frac{24.6}{85.9}=0.29$

At 37.8 mi/h were $\frac{45.1}{85.9}=0.52$

Now multiplying the speed by the portion and summing them we can have the average velocity:

$57*0.19+44*0.29+37.8*0.52=43.25$

The average velocity the entire trip was 43.25 mi/h

5 0

3 years ago

Let a = {1, 2, 3, 4, 5} b = {1, 3, 5} c = {4, 6}. find the cardinality of the given set. 33. n(a) 34. n(b) 35. n(a ⋃

bearhunter [10]

1) n(a) = 5

The cardinality of a set is the number of elements in that set. So, you simply need to count how many elements belong to the set, and the answer is 5

2) n(b) = 3

Simlarly, in this case you need to count how many elements belong to the set, the answer is 3

3) n(a ⋃ c) = 6

The union of two set is a set composed by all the elements belonging to either one of the sets, with no repetitions. So, since the first set contains all numbers from 1 to 5, the second set contains a 4, which however already belonged to the first set, but also a 6, which is added in the union. So, a ⋃ c = {1,2,3,4,5,6}, and thus n(a ⋃ c)=6.

4) n(a ⋂ c) = 1

The intersection of two sets is a set composed by the elements belonging to both sets. So, 1,2,3 and 5 don't belong to the intersection, because they belong to the first set alone, while 6 doesn't belong to the intersection because they don't belong to the first one. So, a ⋂ c = {4}, and thus n(a ⋃ c)=1.

4 0

4 years ago

Read 2 more answers

A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c

olchik [2.2K]

The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in $E^{c}$ . Find $P(E^{c})$ .

Solution:

Part 1:

$E^{c}$ means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

$E^{c}=S-E \\ \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\ \\ 
E^{c}=(4,5,6)$

Thus the set compliment of E will contain the elements {4,5,6}.So

$E^{c}$ = {4,5,6}

Part 2)

$P(E^{c})$ means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

$P(E^{c})$ = (Number of Elements in $E^{c}$ )/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set $E^{c}$ = $n(E^{c})$ =3

So,

$P(E^{c})= \frac{n(E^{c}) }{n(S)} \\ \\ 
P(E^{c})= \frac{3}{12} \\ \\ 
P(E^{c})= \frac{1}{4}$

7 0

3 years ago