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Mila [183]
3 years ago
7

If sin tetha=root3/2, what is cos tetha?​

Mathematics
1 answer:
Minchanka [31]3 years ago
6 0

Answer:

The value of cos Ф is ± \frac{1}{2}

Step-by-step explanation:

There are important rules for sin Ф and cos Ф

  • sin²Ф + cos²Ф = 1
  • sin²Ф = 1 - cos²Ф
  • cos²Ф = 1 - sin²Ф

∵ sin Ф = \frac{\sqrt{3}}{2}

∴ sin²Ф = (\frac{\sqrt{3}}{2})^{2}

∴ sin²Ф = \frac{3}{4}

→ By using the third rule above

∵ cos²Ф = 1 - sin²Ф

∴ cos²Ф = 1 - \frac{3}{4}

∴ cos²Ф = \frac{1}{4}

→ Take square root for both sides

∴ cos Ф = ± \frac{1}{2}

∴ The value of cos Ф is ± \frac{1}{2}

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Solve the following inequality.<br> 6.37 &lt;-45.36<br> Which graph shows the correct solution?
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Answer:

third number line

Step-by-step explanation:

6.3r < -45.36

r < -7.2

it can't be the first two because those values are 71, 70 etc and we need 7.1, 7.2 etc

and r is less then -7.2

So it has to be the third one

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Answer:

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Step-by-step explanation:

Sam's investment = 828

Paul's investment will be temporarily marked as x.

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We can write in 3 different ways:

828 : 966

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A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se
bearhunter [10]

Answer:

(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63

(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83

And the confidence interval for this case -1.63 \leq \mu_1 -\mu_2 \leq 2.83

Step-by-step explanation:

We know the following info from the problem

\bar X_1 = 14.5 sample mean for the group 1

s_1 = 3.98 the standard deviation for the group 1

n_1= 20 the sample size for group 1

\bar X_2 = 13.9 sample mean for the group 2

s_1 = 4.03 the standard deviation for the group 2

n_2= 17 the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given :

df = n_1 +n_2- 2 = 20+17-2=35

The confidence level is 0.9 or 90% and the significance level is \alpha=1-0.9=0.1 and \alpha/2 = 0.05 and the critical value for this case is:

t_{\alpha/2} = 1.69

And replacing the info given we got:

(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63

(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83

And the confidence interval for this case -1.63 \leq \mu_1 -\mu_2 \leq 2.83

5 0
3 years ago
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