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lana66690 [7]
3 years ago
7

Show that sec^2x+csc^2x/sec^2xcsc^2x

Mathematics
1 answer:
professor190 [17]3 years ago
4 0

Answer:

\frac{1}{csc^{2}(x)}+\frac{1}{sec^{2}(x)}

Step-by-step explanation:

The image below shows step-by-step on how to solve it.

<em>Hope this helps! </em>:)

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Step-by-step explanation:

hoped I helped:)

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Anyone know how to do this
Anton [14]

let

(hypotenuse ) c = 10 in

(perpendicular) a = 8 in

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3 years ago
A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl
Marina86 [1]

Answer:

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 13

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 3.0123

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 3.0123\frac{279.29}{\sqrt{14}} = 224.85

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 664.14 - 224.85 = $439.29

The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

4 0
3 years ago
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