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Each of 6 students reported the number of movies they saw in the past year. Here is what they reported.
8,6,8,5,4,6
Find the mean number of movies that the students saw.
If necessary, round your answer to the nearest tenth.

Answer :

Answer:

[tex]\boxed {\boxed {\sf mean \approx 6.3}}[/tex]

Step-by-step explanation:

We are given this data and asked to find the data.

8, 7, 8, 5, 4, 6

The mean is also called the average. It is found by dividing the sum of the terms by the number of terms.

[tex]mean= \frac {sum \ of \ terms}{number \ of \ terms}[/tex]

First, we calculate the sum of the terms. Add up all the numbers.

  • sum= 8+7+8+5+4+6
  • sum= 38

[tex]mean=\frac{38}{number \ of \ terms}[/tex]

Next, divide by the number of terms. There are 6 terms (also, there are 6 students that reported).

[tex]mean=\frac{38}{6}\\mean=6.33333333333[/tex]

Let's round to the nearest tenth.

  • 6.33333333333

The 3 in the hundredth place tells us to leave the 3 in the tenth place.

[tex]mean \approx 6.3[/tex]

The mean number of movies the students saw is approximately 6.3

Step-by-step explanation:

Terms are

  • 8,6,8,5,4,6

[tex] \rm \mapsto \: mean = \frac{sum \: of \: terms}{number \: of \: terms} \\ \rm \mapsto \: \frac{8 + 6 + 8 + 5 + 4 + 6}{6} \\ \rm \mapsto \: \frac{14 + 13 + 10}{6} \\ \rm \mapsto \: \frac{37}{6} \\ \rm \mapsto \: 6.166 \\ \rm \mapsto \: 6.1 \overline{6} \\ \rm \mapsto \: 6.2[/tex]

[tex]\boxed{\large{\sf Mean\approx 6.2}}[/tex]

[tex]\sf Knowledge\:booster{\begin{cases}\bf{\dag}\:\:\underline{\textsf{Fraction Rules :}}\\\\\bigstar\:\:\sf\dfrac{A}{C} + \dfrac{B}{C} = \dfrac{A+B}{C} \\\\\bigstar\:\:\sf{\dfrac{A}{C} - \dfrac{B}{C} = \dfrac{A-B}{C}}\\\\\bigstar\:\:\sf\dfrac{A}{B} \times \dfrac{C}{D} = \dfrac{AC}{BD}\\\\\bigstar\:\:\sf\dfrac{A}{B} + \dfrac{C}{D} = \dfrac{AD}{BD} + \dfrac{BC}{BD} = \dfrac{AD+BC}{BD} \\\\\bigstar\:\:\sf\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{AD}{BD} - \dfrac{BC}{BD} = \dfrac{AD-BC}{BD}\\\\\bigstar \:\:\sf \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \times \dfrac{D}{C} = \dfrac{AD}{BC}\end{cases}}[/tex]