Which of the following scenarios fits the approach of the scientific method? Label as inductive or deductive reasoning.

1 answer:

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When you look at options b) c) and d) they all have something in common: something happened once, or many times, and from this someone thinks (induces) that this is the rule - this i an example of an inductive reasoning.

The correct answer is a)- the mechanic had a hypothesis which he then tested and rejected - this is how a scientific method works!

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A test consists of 10 true or false questions. To pass the test a student must answer at least eight questions correctly. If the

forsale [732]

<h3>The probability of student passing the quiz with at least 50% of the questions correct is 0.05457.</h3>

Step-by-step explanation:

Here, the total number of T/F question = 10

The minimum answers needed correctly answered = 8

So, student needs to answer at least 8 questions correctly.

Here, the possibility of answering a question correctly = $(\frac{1}{2})$ = p = 0.5

Also, the possibility of answering a question wrong = $(\frac{1}{2})$ = q = 0.5

Now, to pass he needs to answer 8 or more ( 8 , 9 or 10) answers correctly.

P(answering 8 correct answer) = $^{10}C_8(p)^8(q)^2 = ^{10}C_8(0.5)^8(0.5)^2 = 0.0439$

P(answering 9 correct answer) = $^{10}C_9(p)^9(q)^1 = ^{10}C_9(0.5)^9(0.5)^1 = 0.0097$

P(answering 10 correct answer) = $^{10}C_{10}(p)^{10}(q)^0 = ^{10}C_{10}(0.5)^{10}(0.5)^0 = 0.00097$

So, the total Probability = P(8) + P(9) + P(10)

= (0.0439) + (0.0097) + (0.00097)

= 0.05457

Hence, the probability that the student passes the quiz with at least 8 of the questions correct is 0.05457.

6 0

4 years ago

A box with a square base and open top must have a volume of 237276 cm^3. We wish to find the dimensions of the box that minimize

svp [43]

Answer:

$A(x)=\dfrac{x^3+949104}{x}$

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 237276 cubic centimetre. We want to find a formula for the surface area of the box in terms of only x, the length of one side of the square base.

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume,

$V=x^2h=237276\\\\h=\dfrac{237276}{x^2}$