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3 years ago

48 is how many percent of 70

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

3 0

It would be: 48/70 * 100 = 4800/70 = 68.57%

So, your answer is 68.57%

Hope it helped.

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Suppose a random sample of 25 students is selected from a community college where the scores in the final exam (out of 125 point

BartSMP [9]

Answer:

he probability that the sample mean deviates from the population mean = 112 by no more than 4= 0.9050

Step-by-step explanation:

mean = 112

sd = 12/sqrt(25) = 2.4

z = (x - μ)/σ

z = (116 - 112)/2.4 = 1.67

Therefore,

P(X >= 116) = P(z <= (116 - 112)/2.4)

= P(z >= 1.67)

= 1 - 0.9525

= 0.0475

z = (x - μ)/σ

z1 = (108 - 112)/2.4 = -1.67

z2 = (116 - 112)/2.4 = 1.67

Therefore, we get

P(108 <= X <= 116) = P((116 - 112)/2.4) <= z <= (116 - 112)/2.4)

= P(-1.67 <= z <= 1.67) = P(z <= 1.67) - P(z <= -1.67)

= 0.9525 - 0.0475

= 0.9050

6 0

4 years ago

Let G denote the centroid of triangle ABC. If triangle ABG is equilateral with side length 2, then determine the perimeter of tr

zhannawk [14.2K]

We know from Euclidean Geometry and the properties of a centroid that GC=2GM. Now GM=sin60*GA= $\sqrt{3}$ . Hence CM=GC+GM=3* $\sqrt{3} /$ . Now, since GM is normal to AB, we have by the pythagoeran theorem that:
$AC^2=AM^2+GM^2=BM^2+GM^2=BC^2$
Hence, we calculate from this that AC^2= 28, hence AC=2* $\sqrt{7}$ =BC. Thus, the perimeter of the triangle is 2+4* $\sqrt{7}$ .

3 0

4 years ago

Can anyone help me with this? I will mark you as brainliest

creativ13 [48]

Answer:

6×6×6×6×6×6×6=279936

6×6×6×6×6=7776

279936÷7776= 36

36 is the answer but u could express it as: (6²)....Six squared

Step-by-step explanation:

4 0

3 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!

4 0

3 years ago

I need the answer ASAP

Orlov [11]

And that’s what you do

3 0

3 years ago

Read 2 more answers