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3 years ago

Jake's test scores are 86, 70, 72, 100, 96, and 80. What must Jake earn of his last test to earn a mean grade of exactly 85%?

Mathematics

1 answer:

kompoz [17]3 years ago

4 0

Answer:

Step-by-step explanation:

Mean is calculated as

mean = $\frac{sum}{count}$

let x be his test score of his last test , then

$\frac{86+70+72+100+96+80+x}{7}$ = 85 ( multiply both sides by 7 )

504 + x = 595 ( subtract 504 from both sides )

x = 91

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Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

Debora [2.8K]

So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$

Since

$\int\limits^1_0 {x^{2} } \, dx = 13$ ,

Substituting this into the equation the equation, we have

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Learn more about definite integrals here:

brainly.com/question/17074932

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