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Margarita [4]
3 years ago
15

X = –16 or x = –4 x = –10 x = –8 x = 4 or x = 16

Mathematics
1 answer:
OverLord2011 [107]3 years ago
3 0
X is 16 so turn everything into a equation
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given that sin theta= 1/4, 0 is less than theta but less than pi/2, what is the exact value of cos theta
lapo4ka [179]

Answer:

\cos{\theta} = \frac{\sqrt{15}}{4}

Step-by-step explanation:

For any angle \theta, we have that:

(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1

Quadrant:

0 \leq \theta \leq \frac{\pi}{2} means that \theta is in the first quadrant. This means that both the sine and the cosine have positive values.

Find the cosine:

(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1

(\frac{1}{4})^{2} + (\cos{\theta})^{2} = 1

\frac{1}{16} + (\cos{\theta})^{2} = 1

(\cos{\theta})^{2} = 1 - \frac{1}{16}

(\cos{\theta})^{2} = \frac{16-1}{16}

(\cos{\theta})^{2} = \frac{15}{16}

\cos{\theta} = \pm \sqrt{\frac{15}{16}}

Since the angle is in the first quadrant, the cosine is positive.

\cos{\theta} = \frac{\sqrt{15}}{4}

3 0
3 years ago
If you were solving a system of equations and you came to a statement like 4 = 4, what do you know about the solution to the sys
GuDViN [60]
It is infinite solutions.
6 0
3 years ago
Read 2 more answers
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salantis [7]
Thanks! :)

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5 0
2 years ago
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Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak
crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28

We can now calculate the probabilities:

P(0.47

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2

We can now calculate the probabilities:

P(0.47

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0
3 years ago
Jessica went to the mall on Saturday to buy clothes she spent $848 on a shirt and 408 dollars on pants in total how much money d
amid [387]

Answer:

Step-by-step explanation: the money that Jessica spend on clothing is 1256

3 0
3 years ago
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