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Inessa [10]
2 years ago
7

Find the length of the unknown segment (x) in each figure. Use theorem of secant segment and

Mathematics
1 answer:
Over [174]2 years ago
5 0

Answer:

1. x = 6    2. x = 9   3. x = 3

Step-by-step explanation:

Use <u>Intersecting Secants Theorem</u>:

If two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.

1.  AC and AD are secants that intersect at point A

So, AB x AC = AE x AD

         x x 15 = 6 x (6 + 9)

             15x = 90

                 x = 6

2.  KM and KN are secants that intersect at point K

So,  KL x KM = KO x KN

 10 x (10 + 8) = x x 20

              180 = 20x

                  x = 9

3.  IG and IF are secants that intersect at point I

So,  IH x IG = IJ x IF

       x x 39 = 9 x (9 + 4)

           39x = 117

               x = 3

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1,2,3,4,5,6,10

Step-by-step explanation:

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Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy
Alborosie

Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

df = n-1=75-1 =74

The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

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4. Determine whether f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.
Ierofanga [76]

Answer:

YES! we conclude that f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.

Step-by-step explanation:

Given

Given that the function f(x) and g(x) are inverse functions.

f\left(x\right)\:=\:\frac{1}{3}x\:+\:5

g(x) = 3x - 15

To determine

Let us determine whether f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.

<u>Determining the inverse function of f(x) </u>

A function g is the inverse function of f if for y = f(x), x = g(y)

y=\frac{1}{3}x+5

Replace x with y

x=\frac{1}{3}y+5

Solve for y

y=3x-15

Therefore,

YES! we conclude that f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.

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