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Lady bird [3.3K]

4 years ago

12

Multiply and Simplify cos(theta) sin(theta)(sec (theta) + csc (theta))

1 answer:

fredd [130]4 years ago

3 0

Before I put out my answer, I just want make note that sec() = 1 / cos() and csc () = 1 / sin ()

1) Once you distribute cos()sin() to sec(), the cos() will cancel out with the sec() to make it just sin(), and once cos()sin() is distributed to csc(), the sin() will cancel out with the cos()

2) After doing that you should just be left with sin() + cos()

3) And since no more simplification can be done, that is your answer :D

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Write each rational number as a decimal. A)17/40 B)11/24

scZoUnD [109]

A)17/40 = 0.425

B)11/24 = 0.458

4 0

4 years ago

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5x + 4 = -6<br> please step by step explanation

Setler79 [48]

Answer:

x=-2

Step-by-step explanation:

5x+4=-6 so basically what you want to do is try to get x by itself

step 1 subtract 4 from each side

5x=-10

step 2 divide each side by 5

x=-2

3 0

3 years ago

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A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram

azamat

Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The standard deviation of sample mean is inversely proportional to the sample size, <em>n</em>.

So, if <em>n</em> increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of <em>n</em> from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for <em>n</em> = 64 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7$

The standard deviation of the sample mean for <em>n</em> = 196 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4$

The standard deviation of the sample mean decreased from 0.7 to 0.4 when <em>n</em> is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of <em>n</em> from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for <em>n</em> = 784 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2$

The standard deviation of the sample mean for <em>n</em> = 49 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8$

The standard deviation of the sample mean increased from 0.2 to 0.8 when <em>n</em> is decreased from 784 to 49.

Hence, the variance also increases.

6 0

3 years ago

Find the projection of v on to w if v = ⟨-5,-2⟩ and w = ⟨1,-1⟩

eduard

Answer:

Step-by-step explanation:

$\frac{\vec v \bullet \vec w}{||\vec w||}\bullet \vec w$

$\vec v \bullet \vec w = \bullet = -5+2 = -3$

$||\vec w ||=\sqrt{2}$

Hence, the result will be

$\frac{-3\sqrt{2} }{2}$

you can put the scalar in to get

8 0

3 years ago

One million people have equal chances of being called by telephone from a broadcasting show. Determine the probability of one pa

Naddik [55]

One over a million is the probability.

5 0

3 years ago