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3 years ago

Solve for 2. Round to the nearest tenth, if necessary.

Mathematics

1 answer:

inysia [295]3 years ago

6 0

Answer:

Step-by-step explanation:

The reference angle is given as 63 degrees. The sides in question, x and 1, are adjacent to and opposite of this angle, respectively. The trig ratio that utilizes the sides adjacent to and opposite of angles is the tangent ratio; namely:

$tan(63)=\frac{1}{x}$ and rearrange algebraically to get

$x=\frac{1}{tan(63)}$ to get

x = .5

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Write and simplify the polynomial represented by the model 3 squares; 2 black rectangles; 3 small boxes ; 1 black square 6 small

irina1246 [14]

Ansqwewer:

weqweqewqeqweqweqw

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qewqewwqwqeewqeqweqw

4 0

4 years ago

Solve for x on the domain 0 < x < 12:<br> cos x= -0.814

Morgarella [4.7K]

Answer:

Replace 10.0 with 10.04

Step-by-step explanation:

$\cos(x) = - 0.814$

$\cos {}^{ - 1} ( - 0.814) = x$

$x = 2.52$

Next, use the identity

$\cos(x) = \cos(2\pi - x)$

$\cos(2.52) = \cos(2\pi - 2.52)$

$\cos(2.52) = \cos(6.28 - 2.52)$

$\cos(2.52) = \cos(3.76)$

So another solution is 3.76

After that, use the identity

$\cos(x + 2\pi) = \cos(x)$

So use that for 2.52 and 3.76

$\cos(2.52 + 2\pi) = \cos(8.80)$

8.80 is one solutiin

$\cos(3.76 + 2\pi) = \cos(10.04)$

4 0

2 years ago

A book claims that more hockey players are born in January through March than in October through December. The following data sh

astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

3 0

4 years ago

What is the area of a triangle that has a base of 15 1/4in. and a height of 18in?show work

Sauron [17]

Answer : 137.25

Step 1 - Convert the base of a triangle to decimal form

15 1/4 = 15.25

Step 2 - Multiply base by height

18 * 15.25 = 274.5

Step 3 - Divide 274.5 by 2

274.5 ÷ 2 = 137.25

3 0

3 years ago

How to write 7+2-3-8 in sigma notation

Valentin [98]

(7+2-3-8)=0
N=7
Brainliest pts
Hope this helps

7 0

4 years ago