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2 years ago

You can compare irrational numbers using rational approximation.

Mathematics

1 answer:

Natasha_Volkova [10]2 years ago

3 0

Yes you can!!!!!!!!!!!

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The advertising department of a nationally circulated magazine wishes to estimate the mean age of its subscribers within 0.5 yea

zhenek [66]

Answer: Option 'd' is correct.

Step-by-step explanation:

Since we have given that

Margin of error = 0.5 years

Standard deviation = 5

At 90% confidence, z = 1.645

We need to find the minimum size of the sample they must take.

So,

$n=(\dfrac{z\sigma}{E})^2\\\\n=(\dfrac{1.645\times 5}{0.5})^2\\\\n=270.6\\\\n\approx 271$

Hence, Option 'd' is correct.

5 0

2 years ago

6. Solve the problem.

Assoli18 [71]

Answer: 32

Step-by-step explanation:

we have in this case right triangle with a leg equal to 24 and a hypotenuse equal to 40

c^2=a^2+b^2

40^2=24^2+b^2

b^2=1600-576=1024

b=sqrt1024=32

4 0

3 years ago

30 + (1/6)^-2 - (-52) What is the value?<br> Is 62 correct?

slava [35]

Answer:

No, the value is 118

Step-by-step explanation:

30+( 1/ 6 )^−2−(−52)

= 30+36−(−52)

= 66−(−52)

= 118

7 0

2 years ago

Find g(4) and f(g(4)) <br> f(x)=x^2-x+1 <br> g(x)= 3x-5

Ira Lisetskai [31]

Answer:

g(4) = 7 and f(g(4)) = 43

Step-by-step explanation:

First find g:

g(x) = 3x-5

g(x) = 3(4) - 5

g(x) = 12 - 5

g(x) = 7

Plug in:

f(g(4))

f(7)

Now, find f:

7^2 - 7 + 1

49 - 7 + 1

42 + 1

f(x) = 43

5 0

2 years ago

) Find the coordinates of the point A' which is the symmetric point

soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

Also, product of slope will be -1 .( Since, they are parallel )

$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

x = 2y - 1

So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

Also, C satisfy given line :

$2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}$

Also,

$x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}$

Therefore, the symmetric points is $A'(\dfrac{31}{5}, \dfrac{18}{5})$ .

7 0

2 years ago