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2 years ago

12

The number of students who signed a petition for more after-school sports is 195, or 30% of the students at school. Terry's

calculations show that there are 58.5 students, but he knows that number is incorrect. What is the correct enrollment? What was Terry's error?
Question content area bottom
Part 1
The correct enrollment is

enter your response here student(s). Terry found
▼
the percent of 195 that 30 is
30% of 195
what 195 is 30% of
instead of finding
▼

1 answer:

Sergeu [11.5K]2 years ago

3 0

Answer:

650

Step-by-step explanation:

195

_. of 30

100

and these 30 zero cut and 100 zero also cut and 3 * 195= 65 and 65 x 10 =650

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The area of a playground is 204 yd^2 the width of thr playground is 5 yd longer that its length what the length and the width of

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Area= LW
204 square yards= LW
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4 years ago

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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4 years ago