Find the green shaded areas.

What number has the same value as 28 tens

Digiron [165]

You would multiply 28 times 10 which equals 2,800

4 0

3 years ago

The 65 students in a classical music lecture class were polled, with the following results: 37 like Wolfgang Amadeus Mozart 36 l

Anna35 [415]

Answer:

a) 25

b) 30

c) 10

d) Not Mozart, 6

e) 2

Step-by-step explanation:

We use a Venn Diagram to solve this question.

I am going to say that:

A are the students who like Mozart.

B are the students who like Beethoven

C are the students who like Haydn.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those who only like Mozart, $(A \cap B)$ are those who like Mozart and Beethoven, $(A \cap C)$ are those who like Mozart and Haydn and $(A \cap B \cap C)$ are those who like all three of them.

By the same logic, we have that:

$B = b + (A \cap B) + (B \cap C) + (A \cap B \cap C)$

$C = c + (B \cap C) + (A \cap C) + (A \cap B \cap C)$

We start finding these values from the intersection:

8 like all three composers

This means that $A \cap B \cap C = 8$

14 like Beethoven and Haydn

This means that:

$(B \cap C) + (A \cap B \cap C) = 14$

So

$B \cap C = 6$

21 like Mozart and Haydn

This means that:

$(A \cap C) + (A \cap B \cap C) = 21$

Then

$A \cap C = 13$

14 like Mozart and Beethoven

This means that:

$(A \cap B) + (A \cap B \cap C) = 14$

$A \cap B = 6$

31 like Franz Joseph Haydn

This means that C = 31. So

$C = c + (B \cap C) + (A \cap C) + (A \cap B \cap C)$

$31 = c + 6 + 13 + 8$

$c = 4$

36 like Ludwig van Beethoven

This means that $B = 36$

So

$B = b + (A \cap B) + (B \cap C) + (A \cap B \cap C)$

$36 = b + 6 + 6 + 8$

$b = 16$

37 like Wolfgang Amadeus Mozart

This means that A = 37. Then

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$37 = a + 6 + 13 + 8$

$a = 10$

a. exactly two of these composers?

$(A \cap B) + (A \cap C) + (B \cap C) = 6 + 13 + 6 = 25$

b. exactly one of these composers?

$a + b + c = 10 + 16 + 4 = 30$

c. like only Mozart?

$a = 10$

d. like Beethoven and Haydn, but not Beethoven?

I will use not Mozart.

So $B \cap C = 6$

Not Mozart, 6.

e. like none of these composers?

At least 1:

$(A \cup B \cup C) = a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 10 + 16 + 4 + 6 + 13 + 6 + 8 = 63$

The total is 65

So 65 - 63 = 2 like none of these composers

4 0

3 years ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)