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Sunny_sXe [5.5K]
1 year ago
8

In a certain school 70% of the students in first-year chemistry have had algebra. If there are 290 students in first-year chemis

try, how many of them have had algebra?
Mathematics
1 answer:
KiRa [710]1 year ago
8 0

The number of the students in the school that have had algebra is 203

<h3>How to determine the number of students?</h3>

The given parameters in the question are

  • Proportion of students taking algebra, p = 70%
  • Number of students in the school, n = 290

The number of students that had algebra in the school is calculated as:

Algebra = Proportion of students taking algebra * Number of students in the school

The above equation can be represented as:

Algebra = np

Substitute values for n and p in the above equation

So, we have

Algebra = 290 * 70%

Express 70% as decimal i.e. 0.70

So, we have

Algebra = 290 * 0.70

Evaluate the product i.e. multiply 290 and 0.70

So, we have

Algebra = 203

Hence, the number of the students in the school that have had algebra is 203

Read more about expected values at:

brainly.com/question/15858152

#SPJ1

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HELPPP!!!! Both questions are incredibly confusing to me !!!!!
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Step-by-step explanation:

I'm going to start with problem 3. You need to become familar with the kind of tricks teachers play on you.

Problem 3 depends on getting RS = RW.

RT = RT                                    Reflexive property

<STR = <WTR                           A straight line having 1 rt angle  actually has 2

ST = TU                                    They are marked as equal

Triangle STR=Triangle WTR    SAS

RS = RW                                    Corresponding parts of = triangles are =

8x = 6x + 5                                Subtract 6x from both sides

8x -6x = 6x - 6x + 5                   Combine

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2x/2 = 5/2

x = 2.5

RU = 6*2.5 + 5

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Now we can play with Question 4.

This question depends on the method used in three, although not entirely.

What you need to know is that W is on RT when you take a ruler and make RT longer. You can put W anywhere as long as it is on RT when it is made longer.

Directions

Make RT longer. Draw down and to your right.

Put a point anywhere on the length starting at T. Label this new point as W. There's your W. It goes anywhere on the part of RT made longer.

Draw UW.

Label UW as 8

Now draw another line from S to W. Guess what? By the methods used in question 3, it's also 8. So SW = 8

TW = TW                        Reflexive

<UTW = STW                 Same reason as in 3. UtW is a right angle

UT = ST                          Given (the marking tells you so.

ΔUTW = ΔSTW              SAS

UW = SW                       Corresponding parts of = triangles are =

SW  = 8

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May I have some help?
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