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elixir [45]
3 years ago
10

Item 32 On three 150-point geography tests, you earned grades of 88%, 94%, and 90%. The final test is worth 250 points. What per

cent do you need on the final to earn 93% of the total points on all tests?
PLEASE HELPPP!!!!!!!!!
Mathematics
1 answer:
stira [4]3 years ago
3 0
97.2% on the final.
Find the total amount of points needed to get a 93%. 150+150+150+250 = 700 total points. 700*.93 = 651 total points needed to get 93%. find the points received from the other quizzes and subtract the sum from 651 to find how many points need to be scored on the final.
150*.88=132 (test 1)
150*.94=141 (test 2)
150*.90=135 (test 3)
132+141+135= 408

651-408 = 243 points to score on the final exam.
243/250 =.972 or 97.2%
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Which expression is equivalent to *picture attached*
DiKsa [7]

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The correct option is;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right )

Step-by-step explanation:

The given expression is presented as follows;

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right )

Which can be expanded into the following form;

\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3  \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left  n^2 + 3  \times\sum\limits _{n = 1}^{50}  n

From which we have;

\sum\limits _{k = 1}^{n} \left  k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}

\sum\limits _{k = 1}^{n} \left  k = \dfrac{n \times (n+1) }{2}

Therefore, substituting the value of n = 50 we have;

\sum\limits _{n = 1}^{50} \left  k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}

\sum\limits _{k = 1}^{50} \left  k = \dfrac{50 \times (50+1) }{2}

Which gives;

4 \times \sum\limits _{n = 1}^{50} \left  n^2 =  4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}

3  \times\sum\limits _{n = 1}^{50}  n = 3  \times \dfrac{n \times (n+1) }{2} = 3  \times \dfrac{50 \times (51) }{2}

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3  \times \dfrac{50 \times (51) }{2}

Therefore, we have;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right ).

4 0
3 years ago
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