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natita [175]
3 years ago
15

The commutative property does not apply to which of the following operations?

Mathematics
1 answer:
Ulleksa [173]3 years ago
4 0
The answer is division and subtraction. I hope you get it right
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Which of the following best describes the relationship between the variables on the scatter plot below?
lesya [120]

Answer:

Step-by-step explanation:

C Linear association

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Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
1 year ago
I need help with a math question ASAP!! please Below is a screenshot of the problem
Nady [450]

Answer:

Pretty sure it's B, the second one from the top

3 0
3 years ago
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Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. Thi
Dovator [93]

Complete question:

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.

A) Find the mean and standard deviation for the number of girls in groups of 36 births.

B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.

C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?

Answer:

a) mean = 18

Standard deviation =3

b) low range = 12

High range = 24

c) The result of 33 girls is significantly high. Yes, the method is effective.

Step-by-step explanation:

Given:

p = 0.5

n = 36

a) The mean is the product of n and p

Mean u = np

u = 36 * 0.5 = 18

The standard deviation is the square root of the product of n and p&q.

S.d ó = \sqrt{npq}

= \sqrt{np(1-p)}

= \sqrt{36(0.5)(1-0.5)} = \sqrt{9} = 3

b) To find the range rule of thumb:

• For low range

Low range = u - 2ó

= 18 - (2 * 3)

= 12

• High range

= u + 2ó

= 18 + (2*3)

= 24

c) The result is significantly high, because 33 is greater than 24 girls.

A result of 33 girls would prove the method as effective.

8 0
3 years ago
Plz help me because I need it ​
Musya8 [376]

Answer:

C

Step-by-step explanation:

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