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VLD [36.1K]
3 years ago
8

Which of the following can be useful when finding the value of a variable? Check all that apply. A. drawing a diagram B. writing

an equation C. reading a table D. drawing a graph
Mathematics
2 answers:
andre [41]3 years ago
6 0

Answer:

B. Writing an equation is the most useful

Step-by-step explanation:

We are required to find the best method which can be used to find the value of a variable.

Since, we know,

<em>While writing an equation, we assign the dependent variable in terms of the independent variable.</em>

These equations can be easily solved using the methods such as substitution and elimination.

Thus, they give the exact value of the variables.

While options A, C and D, does not give the value of the variables over the domain.

So, the best option useful to find the value of the variable is to write an equation.

That is, option B is the most useful.

earnstyle [38]3 years ago
5 0
Writing an equation would be most helpful, but depending on the situation drawing a diagram or reading a table could work better.
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8/5 or 1.6 you just need to find the rise over the run
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3 years ago
The teacher said estimate this with out counting 1 by 1 then divide it by 20 HELP PLEASSEE
WARRIOR [948]
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7 0
3 years ago
A certain plant runs three shifts per day. Of all the items produced by the plant, 50% of them are produced on the first shift,
Snezhnost [94]

Answer:

0.2941 = 29.41% probability that it was manufactured during the first shift.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Defective

Event B: Manufactured during the first shift.

Probability of a defective item:

1% of 50%(first shift)

2% of 30%(second shift)

3% of 20%(third shift).

So

P(A) = 0.01*0.5 + 0.02*0.3 + 0.03*0.2 = 0.017

Probability of a defective item being produced on the first shift:

1% of 50%. So

P(A \cap B) = 0.01*0.5 = 0.005

What is the probability that it was manufactured during the first shift?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.005}{0.017} = 0.2941

0.2941 = 29.41% probability that it was manufactured during the first shift.

6 0
3 years ago
Find the common ratio and the next term for the sequence shown. Express your answers as fractions or as decimals.
Cerrena [4.2K]
The common ratio of the sequence is calculated by dividing the second term by the first term or dividing the third term by the second term. That is,
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or                            r = 187 / 250 = 0.748 = 0.75
To get the next term, we multiply 187 by 0.75 giving us an answer of 140.25 or 141. 
7 0
3 years ago
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Answer:

19 or -28

Step-by-step explanation:

is it 24 - 5 or 24 - 52?

4 0
2 years ago
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