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SVETLANKA909090 [29]
3 years ago
14

A small school has three foreign language classes, one in French, one in Spanish, and one in German. How many of the 34 students

enrolled in the Spanish class are also enrolled in the French class?(1) There are 27 students enrolled in the French class,and 49 students enrolled in either the French class,the Spanish class,or both of these classes.
(2) One half of the students enrolled in the Spanish class are enrolled in more than one foreign language class.
Mathematics
1 answer:
serg [7]3 years ago
5 0

Answer:

Statement 1 is sufficient

And the solution is 12

Statement 2 is insufficient

Step-by-step explanation:

Statement 1 : There are 27 students enrolled in the French class,and 49 students enrolled in either the French class,the Spanish class,or both of these classes.

Let S = number of Spanish-only students

F = number of French-only students

B = number of students taking both, which is the number the prompt question wants.

Given that;

S + B = 34

F + B = 27

S + F + B = 49

The equations above are from the question and statement 1.

Solving for B, we have;

(S + B) + (F + B) = 34 + 27

B + S + B + F = 61

S + B + F = 49

B + 49 = 61

B = 61 - 49 = 12

Therefore statement 1 is sufficient to give a specific and numerical solution to the question.

Statement 2: One half of the students enrolled in the Spanish class are enrolled in more than one foreign language class.

The statement 2 states that half of the students enrolled in the Spanish class are also enrolled in one foreign language but did not specify which proportion or fraction of them are enrolled in german or french therefore a specific numerical solution cannot be derived from this statement.

Hence, statement 2 is insufficient.

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sarah can complete a project in 90 minutes and her sister betty can complete it in 120 minutes if they both work on the project
alukav5142 [94]

Answer:

It will take them approximately 51.43 minutes to complete the project together

Step-by-step explanation:

This is what is called a "shared job" problem.

The best way to work on them is to start by finding the "portion" of the job done by each of the people in the unit of time.

So, for example, Sarah completes the project in 90 minutes, so in the unit of time (that is 1 minute) she completed 1/90 of the total project

Betty completes the project in 120 minutes, so in the unit of time (1 minute) she completes 1/120 of the total project.

We don't know how long it would take for them to complete the project when working together, so we call that time "x" (our unknown).

Now, when they work together completing the entire job in x minutes, in the unit of time they would have done 1/x of the total project.

In the unite of time, the fraction of the job done together (1/x) should equal the fraction of the job done by Sarah (1/90) plus the fraction of the job done by Betty. This in mathematical form becomes:

\frac{1}{x} =\frac{1}{90} +\frac{1}{120}\\\frac{1}{x} =\frac{4}{360} +\frac{3}{360}\\\frac{1}{x} =\frac{7}{360} \\x=\frac{360}{7} \\x=51.43\,\,min

So it will take them approximately 51.43 minutes to complete the project together.

8 0
3 years ago
What is the answer to 36÷9 using a number line?
Katarina [22]

To divide 36/9 using the number line you have to jump from zero with length of 9 until reach 36, and the result will be the number of jumps.

I do the jumps by steps, but you can draw in the number line:

0. First jump from 0 to 9.

,

1. Second jump from 9 to 9+9=18.

,

2. Third jump from 18 to 18+9=27.

,

3. Fourth jump from 27 to 27+9=36.

,

4. Great!! We already reach 36.

So, we need four jumps of 9 to reach 36 from 0.

So, the result is 36/9=4

4 0
1 year ago
A lock on a bank vault consists of 3 dials, each with 30 positions. In order for the vault to open, each of the three dials must
IRISSAK [1]
30 x 30 x 30 = 27000
7 0
3 years ago
I need help with this please!
Gennadij [26K]

Answer:

5ft

Step-by-step explanation:

perimeter=l+l+w+w

l=12

p=24+2w

34-24=10

10/2=5

8 0
2 years ago
Two mechanics worked on a car. the first mechanic worked for 10 hours, and the second mechanic worked for 15 hours. together the
Yuliya22 [10]
Let x and y be the rate per hour for Mechanic 1 and 2 respectively.

Then,
x+y=205 ---------- (1)
Additionally,
10x+15y = 2500 ------ (2)

Solving equations 1 and 2;
Multiply (1) by 10 and subtract two from the result,
10x+10y = 2050
10x +15 y =2500
----------------------
        -5 y = --450 => y = 450/5 = $90
Then,
x= 205-90 = $115

Therefore,
Mechanic 1 charged $115 per hour
Mechanic 2 charged $90 per hour
3 0
3 years ago
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