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2 years ago

15

The circumference of a woman’s basketball must be between 28.5 in. and 29 in., inclusive. What is a compound inequality that cou

ld show this?

2 answers:

Stolb23 [73]2 years ago

8 0

Answer:

The ball should be BOTH greater than 28.5 AND less than 29 (although I don't think it's in inches...) Since it is not explicitly stated, let's assume we're talking about the basketballs diameter, D.

So now this may be written D > 28.5 and D < 29. Combining these two, we can write 28.5 < D < 29.

Also not explicitly stated is whether or not the basketball may be equal to 28.5 or 29. If, indeed, that is the case, then instead we may write this as D >= 28.5 and D <= 29, or combining, 28.5 <= D <= 29.

Hope this helps :D.

Paladinen [302]2 years ago

6 0

Answer:

Greater than or equal to

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A class survey found that 29 students watched television on Monday, 24 on Tuesday, and 25 on Wednesday. Of those who watched TV

gizmo_the_mogwai [7]

Answer:

There were 49 students in the class

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the students that watched TV on Monday

-The set B represents the student that watched TV on Tuesday.

-The set C represents the students that watched TV on Wednesday.

We have that:

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

In which a is the number of students that only watched TV on Monday, $A \cap B$ is the number of adults that watched TV both on Monday and Tuesday, $A \cap C$ is the number of students that watched TV both on Monday and Wednesday, and $A \cap B \cap C$ is the number of students that watched TV on every day.

By the same logic, we have:

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

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

This diagram has the following subsets:

$a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

The sums of all of this values is the number of student that were there in the class. This means that we want to find the value of T:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = T$

We start finding the values from the intersection of three sets.

Solution:

12 students watched TV on all three days:

$A \cap B \cap C = 12$

14 students watched TV on both Monday and Tuesday

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

$A \cap B = 14 - 12$

$A \cap B = 2$

Of those who watched TV on only one of these days, 13 choose Monday, 9 chose Tuesday, and 10 chose Wednesday.

$a = 13, b = 9, c = 10$

29 students watched television on Monday:

$A = 29$

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

$29 = 13 + 2 + (A \cap C) + 12$

$A \cap C = 29 - 27$

$A \cap C = 2$

24 on Tuesday

$B = 24$

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

$24 = 9 + (B \cap C) + 2 + 12$

$B \cap C = 24 - 23$

$B \cap C = 1$

Now we have every value needed to find T:

$T = a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$T = 13 + 9 + 10 + 2 + 2 + 1 + 12$

$T = 49$

There were 49 students in the class

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