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Novosadov [1.4K]
3 years ago
12

Use the value of the trigonometric function to evaluate the indicated function.

Mathematics
1 answer:
Sonbull [250]3 years ago
6 0

Answer:

\displaystyle \sin t = - \frac{5}{6}

Step-by-step explanation:

Recall that since sine is an odd function:

\displaystyle \sin (-\theta) = -\sin \theta

Hence:

\displaystyle \sin(-t) = -\sin t = \frac{5}{6}

Therefore:

\displaystyle \sin t = - \frac{5}{6}

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Weight of a rock: In a geology course, students are learning to use a balance scale to accurately weigh rocks. One student plans
statuscvo [17]

Answer:

The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.

Step-by-step explanation:

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.  

The confidence interval is given by

CI = \bar{x} + t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\

Where \bar{x} is the mean weight \sigma is the standard deviation t_{\alpha/2} is the critical value from t-table and n is the sample size.

The term t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) is known as margin of error.

As the sample size is decreased the corresponding margin of error increases which results in wider confidence interval which means smaller precision.

The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.

We can say with 95% confidence that the true mean weight of the rock is within the interval of (25.2, 29.1).

7 0
3 years ago
PLEASE SOMEONE HELP ME PLEASE :(
creativ13 [48]

Answer:

1 and 5= corresponding angles.

4 and 2=vertical opposite angle

3 and 6=co interior angle.

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5 and 3= alternate interior angle

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3 0
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Read 2 more answers
The half-life of cesium-137 is 30 years. Suppose we have a
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(a) If y(t) is the mass (in mg) remaining after t years, then y(t) = y(0) e^{kt}=100e^{kt}.

y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}

(b) y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}

(c)
100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}
 
3 0
3 years ago
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