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Nataly_w [17]
2 years ago
10

Grace was given the description “three less than the quotient of a number squared and nine, increased by eight” and was asked to

evaluate it when n = 6. Her work is shown below.
Step 1: mc021-1.jpg
Step 2: mc021-2.jpg
Step 3: mc021-3.jpg
Step 4: mc021-4.jpg
Step 5: 7

In which step did she make an error?
step 1
step 2
step 4
step 5
Mathematics
1 answer:
NeTakaya2 years ago
7 0
Given that the steps are not shown, I am going to show you all the rigtht steps in detail and so you will be able to compare and tell which step from you list contains an error:

1) Traslation into algebraic language:

<span>“three less than the quotient of a number squared and nine, increased by eight”

3 - (n^2) / 9

2) Evaluate it when n = 6 (replace the value of n)

3 - (3^2) / 9

3) solve the parenthesis

3 - 9/9

4) divide 9/9

3 - 1

5) Subtract

2
</span>
Now you know all the steps and can tell which of the steps on the list is wrong.
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What's 3/4 times 7/9 in simplest form
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3/4 times 7/9:

First, we must use the rule for multiplying fraction. The rule is: a/b times c/d = ac/ bd. Let's substitute our problem into that formation.
\frac{3 \times 7}{4 \times 9}

Second, we can now simplify. (3 times 7 = 21) and (4 times 9 = 36). 
\frac{21}{36}

Third, since our fraction is not in the simplest form, we can simplify it down by listing the factors of both the numerator and denominator and then finding the greatest common factor (GCF). 

Factors of 21: 1, 3, 7, 21
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

The GCF is 3, since only 1 and 3 were the common factors and since 3 is the greatest common factor. 

Fourth, now we can divide our numerator (21) and denominator (36) by our recently found GCF which was 3. 

21 \div 3 = 7 \\ 36 \div 3 = 12

Our new fraction in its simplest form is 7/12.

Answer in fraction form: \fbox {7/12}
Answer in decimal form: \fbox {0.5833}
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Mathew rolls a number cube labeled with the numbers 1−6. He then flips a fair coin. What is the probability that he rolls a 4 an
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Answer: 0.083


Step-by-step explanation:

Numbers on cube=6

faces on coin=2

Therefore, the total outcomes=6\times2=12

Now, the favorable outcome that he rolls a 4 and flips a head=1

The probability that he rolls a 4 and flips a head=\frac{\text{favorable outcome}}{\text{total outcome}}

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A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting
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Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= \frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees}  = \frac{6}{15}

b.) The probability that a secretary wins a ticket on second draw.  It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant  wins on first draw)

=\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}

= \frac{\frac{5}{15}  \times \frac{6}{14}}{\frac{5}{15} }  = \frac{6}{14}  .

The probability that  a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw  = \frac{6}{15} is not true.

The probability that a secretary wins on the second draw  = \frac{number \hspace{0.1cm} of  \hspace{0.1cm} secretaries  \hspace{0.1cm} remaining } { number  \hspace{0.1cm} of  \hspace{0.1cm} employees  \hspace{0.1cm} remaining}  = \frac{6 - 1}{15 - 1}  = \frac{5}{14}

c.) The probability that a consultant wins on the first draw  =

\frac{number \hspace{0.1cm} of  \hspace{0.1cm} consultants  \hspace{0.1cm}  } { number  \hspace{0.1cm} of  \hspace{0.1cm} employees  \hspace{0.1cm} }  = \frac{5 }{15}  = \frac{1}{3}

d.) The probability of two secretaries winning both tickets

= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)

= \frac{6}{15}  \times \frac{5}{14}  = \frac{1}{7}

Therefore the only statement that is not true is b.)

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