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dmitriy555 [2]
3 years ago
11

Find the mean set of this data set 25,45,12,34,3645,19,20

Mathematics
2 answers:
alisha [4.7K]3 years ago
6 0
Mean=( 25+45+12+34+3645+19+20)/7=542.8571429=543 (3sf)
Dimas [21]3 years ago
5 0
25+45+12+34+36+45+19+20/8= 29.5
Therefore 29.5 is your mean. (:
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Read 2 more answers
Evaluate the following integral using trigonometric substitution.
wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function \int\limits {\sqrt{a^{2} -x^{2} } } \, dx, using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as asin \theta i.e x = a sin\theta.

All integrals in the form \int\limits {\sqrt{a^{2} -x^{2} } } \, dx are always evaluated using the substitute given where 'a' is any constant.

From the given integral, \int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx where a = 7 in this case.

The substitute will therefore be   x = 7 sin\theta

2.) Given x = 7 sin\theta

\frac{dx}{d \theta} = 7cos \theta

cross multiplying

dx = 7cos\theta d\theta

3.) Rewriting the given integral using the substiution will result into;

\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta  } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)}   } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)}   }}} \, 7cos\theta d\theta\\

= \int\limits343 cos^{2}  \theta \, d\theta

8 0
4 years ago
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