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3 years ago

The sine and cosine of one of the acute angles in a right triangle are equal. What is the tangent of the angle?

Mathematics

1 answer:

Gnoma [55]3 years ago

4 0

Answer:

Step-by-step explanation:

Since tan(x) = sin(x)/cos(x), if sin(x) = cos(x) then tan(x) must be 1.

You can even reason that the acute angle is 45° and sin(x)=cos(x) = $\frac12\sqrt2$ .

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Suppose there are 310 first-year lawyers in a particular metropolitan area with an average starting salary of $156,000 and a sta

vlada-n [284]

Answer:

$ 2263

Step-by-step explanation:

In this case to calculate the standard error of the mean, we only need the standard deviation (sd) and the number of the random sample (n).

sd = 13000

n = 33

SE = sd / (n ^ (1/2))

replacing:

SE = 13000 / (33 ^ (1/2))

SE = 2263.01

What the standard error of the mean for a random sample of 33 first-year lawyers means is $ 2263

6 0

3 years ago

What is the range of the data shown on the stem-and-leaf plot? <br>93<br> 43 <br>50<br> 48

notsponge [240]

Answer: C) 50

Step-by-step explanation:

The smallest number on the plot is 43. The largest is 93. The range of a chart is the largest number - the smallest number. Thus, simply do 93-43 to get 50.

Hope it helps <3

5 0

2 years ago

Read 2 more answers

The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p

VLD [36.1K]

Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

$\chi^2_{1-\frac{\alpha}{2}}= 51.770$

Chi square critical value for upper tail =

$\chi^2_{\frac{\alpha}{2}}= 81.085$

80% confidence interval:

$\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13$

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0

3 years ago

Can someone help me on this question

matrenka [14]

What's the question??

5 0

2 years ago

A blank CD can hold 70 minutes of music. So far you have burned 25 minutes of music onto the CD. You estimate that each song las

zaharov [31]

Answer:

The possible numbers of songs that you can burn onto the CD is 11.

Step-by-step explanation:

The total storage capacity of CD = 70 minutes

Storage already used = 25 minutes

Hence, The available storage = Total available storage - Used Storage

= 70 minutes - 25 minutes = 45 minutes

Hence, the CD has 45 minutes storage left.

Now, each song takes up 4 minutes of storage.

⇒ $\textrm{Total number of songs possible} = \frac{\textrm{Total storage left}}{\textrm{Time taken by each song}}$

= $\frac{45 minutes}{4minutes } = 11.25$

⇒The number of songs possible = 11.25 ≈ 11

Hence, the possible numbers of songs that you can burn onto the CD is 11.

4 0

3 years ago