Can someone help me with this math homework please!

one interior angle of an hexagon is 170°, and each of the remaining interior angles is equal to y° ,find the value of y.

Inessa [10]

Answer:

y = 110

Step-by-step explanation:

The formula for the sum of the interior angles of a polygon with n sides is: (n-2) x 180 degrees. A hexagon has 6 sides, so if we plug this value into the equation, we get (6-2) x 180 = 4 x 180 = 720 degrees. Since we have one angle which is 170 degrees and the other five angles are the same (y), we can make the equation 170 + 5y = 720. We subtract 170 on both sides and get the equation 5y = 550. Dividing both sides by 5, we get y = 110, which is our answer.

8 0

2 years ago

Find the slope of the graphed line

Contact [7]

Answer:

slope = 3/5

Step-by-step explanation:

slope = rise/run = 3/5

or (longer way w/ the points)

slope = (y2-y1)/(x2-x1) = (5-2)/(4-(-1)) = 3/5

3 0

3 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

Sorry I ain’t know why it look like that buh can someone help asap

GalinKa [24]

Answer:

$for \: \: dmain \: of \: {f}^{ - } \: \: is \: equal \: f \: range \\ y = \sqrt{x } - 5 \\y \: range \: is \: y \geqslant - 5\\ y + 5 = \sqrt{x} \\ x = {(y + 5)}^{2} \\ {f}^{ - 1} = {(x + 5)}^{2} \\ \\ dmain \: of \: \: {f}^{ - 1} \: x \geqslant - 5$

5 0

2 years ago

Trish is solving the equation x + 10 = 8 for the value of x. x + 10 + 10 = 8 + 10 x = 18 (2)x = (2)18 x = 36 What error did Tris

zlopas [31]

X+10+10=8+10 is not equal to x=18. It must be x+20=18 x=-2

5 0

4 years ago

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