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Sedbober [7]
3 years ago
6

A Movie theater sold 5 times as many children's tickets as adult tickets to an afternoon show. If 132 tickets were sold in all,

how many were children's tickets?

Mathematics
1 answer:
Vikentia [17]3 years ago
7 0
22x5=110
110+22=132
so if they sold 22 adult tickets 5 times that many is 110. they sold 110 children tickets, plus the 22 adult tickets equals the total of 132 tickets. So the answer is 110
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We test for a hypothesized difference between two population means: H0: μ1 = μ2. The population standard deviations are unknown
Lina20 [59]

Answer:

The degrees of freedom associated with the critical value is 25.

Step-by-step explanation:

The number of values in the final calculation of a statistic that are free to vary is referred to as the degrees of freedom. That is, it is the number of independent ways by which a dynamic system can move, without disrupting any constraint imposed on it.

The degrees of freedom for the t-distribution is obtained by substituting the values of n1​ and n2​ in the degrees of freedom formula.

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Therefore, the degrees of freedom associated with the critical value is 25.

4 0
3 years ago
What is the slope of y=17x+43
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5 0
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Read 2 more answers
Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))
NNADVOKAT [17]

I believe the given limit is

\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)

Let

a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1

Now rewrite the expression as a difference of cubes:

a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}

Then

a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2

The limit is then equivalent to

\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}

From each remaining cube root expression, remove the cubic terms:

a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}

(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}

b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}

Now that we see each term in the denominator has a factor of <em>x</em> ², we can eliminate it :

\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}

=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}

As <em>x</em> goes to infinity, each of the 1/<em>x</em> ⁿ terms converge to 0, leaving us with the overall limit,

\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}

8 0
3 years ago
Y=8x-3/-4x-10 has a vertical asymptote with equation(enter equation of the vertical asymptote
xz_007 [3.2K]

Answer:

x=-2.5 if the function is f(x)=\frac{8x-3}{-4x-10}

Step-by-step explanation:

y=\frac{8x-3}{-4x-10} has discontinuities when the denominator is 0.

You will either have a hole or a vertical asymptote depending on what happens to the numerator after you find when the bottom is 0.

That is whatever you found that makes the bottom 0, if it makes the top also 0 then you will have a hole at x=the number that made the bottom 0.

If it makes the top anything other than 0, then it is a vertical asymptote at x=the number you found that made the bottom 0.

Let's do this now.

When is -4x-10 equal to 0?

We have to solve the equation:

-4x-10=0

Add 10 on both sides:

-4x=10

Divide both sides by -4:

x=10/-4

Reduce by dividing top and bottom by 2:

x=5/-2

x=-5/2

or

x=-2.5  (if you want decimal form)

Now does it make the top 0? This is the deciding factor on whether you have a hole at x=-2.5 or a vertical asymptote at x=-2.5.

Let's see.

8(-2.5)-3=-23

Since the top is not 0 at x=-2.5 then you have a vertical asymptote at x=-2.5.

If the top were 0, then you would have had a hole at x=-2.5.

5 0
3 years ago
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