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3 years ago

Determine the solution for x^2 + 36 > 12x

Mathematics

1 answer:

Roman55 [17]3 years ago

4 0

X<6 or x>6 hope this helps

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Answer:

y=6^x-2

Step-by-step explanation:

Start with the parent function, a^x. The graph looks like it has been translated b units down, so our function is a^x+b. Now at x=0, y=-2. So b=-2. Next at x=1, y=3. 3=a^(1)-2, a=6. y=6^x-2 is the equation

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4 years ago

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3 years ago

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3 years ago

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3 years ago

HighTech Inc. randomly tests its employees about company policies. Last year in the 560 random tests conducted, 26 employees fai

Tanzania [10]

Answer:

The 98% confidence interval for the proportion of applicants that fail the test is (0.025, 0.067).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

560 random tests conducted, 26 employees failed the test. This means that $n = 560, \pi = \frac{26}{560} = 0.046$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.33$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 - 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.025$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 + 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.067$

The 98% confidence interval for the proportion of applicants that fail the test is (0.025, 0.067).

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4 years ago