The figure below shows a shaded rectangular region inside a large rectangle:

Mathematics

2 answers:

Svetlanka [38]3 years ago

8 0

First we need to find the areas of both rectangles.

And area of larger rectangle = 10*5=50 square units

Area of smaller rectangle = 7*3 = 21 square units

Area between the two rectangles, = 50-21=29 square units

And since, we have to find the probability that a point is choosen is not in the shaded rectangle that is the smaller rectangle, so it means we have to find the probability of the point is choosen in the area between the two rectangles . Therefore,

Required probability =

$=\frac{29}{50}*100 =58%$

Therefore correct option is the second option .

sergiy2304 [10]3 years ago

5 0

Answer:

Step-by-step explanation:

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jeyben [28]

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3 years ago

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