Please help me find the measure of the arc or angle indicated. (Angles on circles)

Mathematics

1 answer:

nata0808 [166]3 years ago

5 0

Answer: 9. 54; 10. 220; 11. 71; 12. 262

Step-by-step explanation: Chord and tangent intersection angle is half the arc measure or circle, and inscribed angles are also half the included arc measure

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A Local group claims that the police issue more than 60 speeding tickets a day in thire area. To prove their point, they randoml

MissTica

Answer:

The group's claim is not supported at 1% significance level.

Step-by-step explanation:

given that a local group claims that the police issue more than 60 speeding tickets a day in thire area

The data they collected for 2 weeks is given below:

70 48 41 68 69 55 70

57 60 83 32 60 72 58

$\bar x = 60.2\\s = 12.9\\n=14$

Std error of mean = $\frac{12.9}{\sqrt{14} } \\=3.4477$

$H_0: \bar x=60\\H_a: \bar x >60$

(Right tailed test at 1% significance level)

Mean difference = 0.10

Test statistic t with df 13 = mean difference/std error

= 0.029

p value = 0.4886

Since p >0.01, we accept null hypothesis

The group's claim is not supported at 1% significance level.

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3 years ago

How to solve radical numerator fractions

Nonamiya [84]

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2 years ago

Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$