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mariarad [96]
3 years ago
5

If the mean of five values is 9.6 and four of the values are 8, 14, 7, and 13, find the fifth value. ​

Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:

The missing value is 6

Step-by-step explanation:

We know that there are 5 values but only 4 are given:

⇒ The given values are: 8, 14, 7, and 13

⇒ Let the fifth value be "x"

We also know the mean which is 9.6

<em><u>Solve for the missing value:</u></em>

⇒ 9.6 (mean) = \frac{8 + 14 + 7 + 13 + x}{5 numbers}

<em>multiply 5 on both sides </em>

⇒ 9.6 x 5 = \frac{8 + 14 + 7 + 13 + x}{5 numbers} x 5

⇒ 48 = 8 + 14 + 7 + 13 + x

<em>add the numbers:</em>

⇒ 8 + 14 + 7 + 13 + x = 42 + x

<em>the equation becomes: </em>

⇒ 48 = 42 + x

<em>Now, subtract 42 on both sides:</em>

⇒ 48 - 42 = 42 - 42 + x

⇒ 6 = x

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A skyscraper casts a shadow 200 feet long. If the angle of elevation of the sun is 49 , then the height of the skyscraper is ___
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Read 2 more answers
Evaluate the limit of
tekilochka [14]

Answer:

Option a is correct -- \frac{9}{2}

Step-by-step explanation:

If we put infinity directly into the expression we get ∞ + ∞ expression. In order to circumvent it we divide both numerator and denominator with the greatest exponent.

\lim_{x \to \infty} \frac{9n^{3}+5n-2}{2n^{3}}

 Divide each term by n³

=\lim_{x \to \infty} \frac{\frac{9n^{3}}{n^{3}}+\frac{5n}{n^{3}} -\frac{2}{n^{3}}}{\frac{2n^{3}}{n^{3}} }

=\lim_{x \to \infty}\frac{9(1)+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2(1)} } \\=\lim_{x \to \infty}\frac{9+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2}} \\\\by putting n = infinity \frac{5}{n^{2}} becomes 0 and\frac{2}{n^{3}} becomes 0 \\=\frac{9+0-0}{2} \\=\frac{9}{2}



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