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Alex777 [14]
4 years ago
5

1) Find the mean for the given sample data. Unless indicated otherwise, round your answer to one more decimal place than is pres

ent in the original data values.
The amount of time (in hours) that Sam studied for an exam on each of the last five days is given below.

1.7 7.7 8.3 1.6 5.1

Find the mean study time

A) 4.96 hr
B) 5.45 hr
C) 4.88 hr
D) 24.40 hr


2) In order to avoid staying after work, a quality control analyst inspects the first 100 items produced that day. Which sampling method did she use?

A) cluster
B) systematic
C) stratified
D) convenience


3) What is the probability that two events, A and B, will occur at the same time?

A) P(A)+P(B)
B) P(A and B)
C) P(A or B)

4) A yogurt shop offers 6 different flavors of frozen yogurt and 12 different toppings. How many choices are possible for a single serving of frozen yogurt with one topping?

A) 72
B) 665,280
C) 144
D) 36


5) Find the mean and range. 26 19 23 39 31 34 23 25

A) mean = 28.7; range = 20
B) mean = 27.5; range = 20
C) mean = 26; range = 21
D) mean = 27.5; range = 21

6) A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function mc011-1.jpg. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

A) 1.13 s; 29.25 ft
B) 1.13 s; 31.5 ft
C) 1.13 s; 69.75 ft
D) 2.25 s; 9 ft

7) Use any method to solve the equation. If necessary, round to the nearest hundredth.

4x^2 - 28 x= -11

A) 6.58, 0.42
B) 79.5, –72.5
C) –0.42, –6.58
D) 13.16, 1.67

8) The cost of textbooks for a school increases with the average class size. Identify the independent and dependent quantity in the situation.


A) cost; number of classes
B) number of classes; average class size
C) cost; average class size
D) average class size; cost

9) Find the distance between points P(8, 2) and Q(3, 8) to the nearest tenth.


A) 14.9
B) 61
C) 11
D) 7.8
Mathematics
2 answers:
docker41 [41]4 years ago
5 0

Answer:

C) 4.88 hr; D) convenience; B) P(A and B); A) 72; B) mean = 27.5, range = 20; no function shown to answer the question; A) 6.58, 0.42; D) average class size, cost; D) 7.8

Step-by-step explanation:

#1) To find the mean, we add all of the numbers and divide by the number of data values:

(1.7+7.7+8.3+1.6+5.1)/5 = 24.4/5 = 1.88

#2) This is a convenience sample because it was easy for the analyst to do; she did not use random sampling or any sort of groups, and she did not choose every kth item.

#3) The probability of two events is P(A and B) by definition.

#4) To find the total number of choices, we multiply the number of flavors by the number of toppings:

6(12) = 72

#5) To find the mean, add all of the data values up and divide by the number of data values:

(26+19+23+39+31+34+23+25)/8 = 220/8 = 27.5

To find the range, subtract the highest and lowest values:

39-19 = 20

#6) There is no function given to answer the question.

#7) Using the quadratic formula,

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\=\frac{--28\pm \sqrt{(-28)^2-4(4)(11)}}{2(4)}\\\\=\frac{28\pm \sqrt{784-176}}{8}\\\\=\frac{28\pm \sqrt{608}}{8}\\\\=\frac{28\pm 24.65766}{8}\\\\=\frac{28+24.65766}{8},\frac{28-24.65766}{8}\\\\=\frac{52.65766}{8},\frac{3.34234}{8}\\\\=6.58,0.42

#8) The independent quantity is the one that causes the dependent to change.  In this case, the average class size causes the cost of books to change; this means the average class size is independent and the cost of books is dependent.

#9) Using the distance formula,

d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\=\sqrt{(8-2)^2+(3-8)^2}\\\\=\sqrt{6^2+(-5)^2}\\\\=\sqrt{36+25}=\sqrt{61}=7.8

Semmy [17]4 years ago
4 0
Alright, so #1 is C, 4.88.
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The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6
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Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Poisson distributed, with a mean of 0.6 trips per year

This means that \mu = 0.6n, in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so \mu = 0.6.

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so \mu = 0.6.

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

P(X < 2) + P(X \geq 2) = 1

We want

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so \mu = 0.6(3) = 1.8.

This is

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653

P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975

P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678

P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so \mu = 0.6(6) = 3.6.

This is P(X = 4). So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

4 0
3 years ago
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