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2 years ago

Please help find the probability that x >25.

Mathematics

1 answer:

bija089 [108]2 years ago

5 0

Answer

The total probability is one:

Total probability of being greater than 25 is 1 because the totals of all values less than 25 is 1.

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Which shape has a larger area: a rectangle that is 7 inches by 34 inch, or a square with side length of 212 inches?

GREYUIT [131]

Answer is square with the side of 212 inches. It's area is 212×212 which is bigger than rectangular area.

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2 years ago

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How do you solve x - 1 + 5x = 23

Roman55 [17]

Answer:

X=4

Step-by-step explanation:

First you add like terms, in this case 5x and x. Then you have 6x-1=23. You then add 1 from both sides so you end up with 6x=24. This then results in X=4

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CAN SOMEONE HELP ME PLEASE ASAP!?

valentinak56 [21]

Answer:

last option

Step-by-step explanation:

b) 1/6 = 16.7 to 1dp

c) 45% = 45/100 = 9/20

d) 7/9 = 77.8% to 1dp

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2 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

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2 years ago

What's a parabola? What's a hyperbola?

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Parabola: is a two-dimensional, mirror-symmetrical curve, which is approximately U-shaped when oriented as shown in the diagram below, but which can be in any orientation in its plane. It fits any of several superficially different mathematical descriptions which can all be proved to define curves of exactly the same shape.

Hyperbola:
In mathematics, a hyperbola (plural hyperbolas or hyperbolae) is a type of smooth curve lying in a plane, defined by its geometric properties or by equations for which it is the solution set. A hyperbola has two pieces, called connected components or branches, that are mirror images of each other and resemble two infinite bows

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