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egoroff_w [7]
1 year ago
5

What basic trigonometric identity would you use to verify that

Mathematics
1 answer:
Dafna1 [17]1 year ago
6 0

Given the equation:

\frac{\sin^2x+\text{cos}^2x}{\cos x}=\sec x

Let's determine the trigonometric identity that you could be used to verify the exquation.

Let's determine the identity:

Apply the trigonometric identity:

\sin ^2x+\cos ^2x=1

\cos x=\frac{1}{\sec x}

Replace cosx for 1/secx

Thus, we have:

\begin{gathered} \frac{\sin^2x+\cos^2x}{\frac{1}{\sec x}} \\  \\ =(\sin ^2x+\cos ^2x)(\sec x) \\ \text{Where:} \\ (\sin ^2x+\cos ^2x)=1 \\  \\ We\text{ have:} \\ (\sin ^2x+\cos ^2x)(\sec x)=1\sec x=\sec x \end{gathered}

The equation is an identity.

Therefore, the trignonometric identity you would use to verify the equation is:

\cos ^2x+\sin ^2x=1

ANSWER:

\cos ^2x+\sin ^2x=1

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A company sells boxes of duck calls (d) for $35 and boxes of turkey calls (t) for $45. they make batches of duck calls that fill
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Answer:

Option A is the correct choice.

Step-by-step explanation:

Let d be the number of boxes of duck calls and t be the number of boxes of turkey calls.

We have been given that a company sells boxes of duck calls for $35 and boxes of turkey calls (t) for $45, so the revenue earned from selling d boxes of duck and t boxes of turkey call will be 35d and 45t respectively.

Further, the company plan to make $300. We can represent this information as:

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We are also told that they make batches of duck calls that fill 6 boxes and batches of turkey calls that fill 8 boxes. the company only has 42 boxes. We can represent this information as:

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3 years ago
Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th
Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

Z = \frac{X - \mu}{\sigma}

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Z = \frac{X - \mu}{s}

Z = \frac{990 - 975}{9.34}

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X = 960

Z = \frac{X - \mu}{s}

Z = \frac{960 - 975}{9.34}

Z = -1.61

Z = -1.61 has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

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