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Dvinal [7]
4 years ago
13

CAN SOMEONE HELP ME WITH THIS PLEASE ??

Mathematics
1 answer:
masya89 [10]4 years ago
3 0
Remark
The very first thing you should do is graph the function, which I have done. It is at the bottom of this answer.

Step One
Where is the hole and what kind is it? This graph has a hole at x = 2. At that point the function is 0/0. A hole is defined as the point created by a zero in one of the factors that is the same in the numerator and denominator. Both = 0.

Step 2
Find the vertical asymptote.
That occurs when denominator alone has a factor that is =  0. In this case when x = 3 the whole equation blows up (or down) and you have a vertical asymptote. The graph is discontinuous at this point. 

Step three 
Find the horizontal Asymptote.
Horizontal assymptotes occur when (in this case) the power of the factors in the denominator exceeds the power of the factors in the numerator. In this case that is true and the horizontal assymptote occurs around the line y = 0.


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Gennadij [26K]

Answer:

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4 years ago
According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.
UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

<em>Let </em>\hat p<em> = sample proportion of Americans who can order a meal in a foreign language</em>

The z-score probability distribution for sample proportion is given by;

          Z = \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( \hat p > 0.50)

  P( \hat p > 0.06) = P( \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } > \frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } } ) = P(Z > 0.85) = 1 - P(Z \leq 0.85)

                                                               = 1 - 0.80234 = <u>0.19766</u>

<em>Now, in the z table the P(Z  </em>\leq <em>x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.</em>

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

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