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adelina 88 [10]
4 years ago
14

Find csc0 and cos0 , where 0 is the angle shown in the figure. Give exact values, not decimal approximations.

Mathematics
1 answer:
solniwko [45]4 years ago
8 0

Answer:

csc(θ) =  1.1662

cos(θ)  = 0.5145

Step-by-step explanation:

I attach the image of your question in the picture below

We know the values of the adjacent and opposite cathetus

tan(θ) = Opposite / Adjacent

tan(θ) = 5/3 = 1.666

θ = tan^-1 (5/3)

θ = 59.0362 °

Now that we have the angle, we can easily find the value of the expressions needed

csc(θ) = 1 / sin(θ)  = 1/ (0.85749) = 1.1662

cos(θ)  = 0.5145

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A school needs to buy new notebook and desktop computers for its computer lab. The notebook computers cost $225 each, and the de
almond37 [142]

Answer:

n=19

d=12

$225n+$250d=

$225(19)+$250(12)=

4275+3000

=7275

Step-by-step explanation:

4 0
2 years ago
CAN SOMEONE HELP ME WITH THIS!!
DIA [1.3K]

Answer:

Number of dogs = 6

Number of burgers = 11

Step-by-step explanation:

Number of dogs = x

Number of burgers = 2x - 1

Cost of a dog = $ 1.50

Cost of 'x' dogs = 1.5x

Cost of a burger = $ 2

Cost of (2x - 1) burger = (2x -1) * 2 = 2x *2 - 1*2

                                    = 4x - 2

Total amount collected  = $ 31

1.5x + 4x - 2 = 31 {add 2 to both sides}

        1.5x + 4x = 31 + 2

      1.5x+ 4x  = 33

               5.5x = 33

                    x = 33/5.5

                   x = 6

Number of dogs  = 6

Number of burgers = 2*6 - 1 = 12 - 1 = 11

4 0
3 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

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.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

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

\mu_{\bar x}=\mu=68

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

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

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4 years ago
Experimental verification of opposite side of parallelograms .
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3 years ago
What types of activities would you incorporate into your marathon training and why
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Running so you can build up both endurance and stamina. It’ll keep you in shape so you can last longer in a marathon. Constant conditioning will do the trick.
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