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Reika [66]
1 year ago
12

The measure of one a cite angle of a right triangle is 6 less than twice the measure of the other acute angle find the measure o

f each acute angle . Help quickly please
Mathematics
1 answer:
jeka57 [31]1 year ago
8 0

Answer:

32°, 58°

Step-by-step explanation:

Let one acute angle measure x.

The other acute angle measures 2x - 6.

The sum of the measures of the acute angles of a right triangle is 90.

x + 2x - 6 = 90

3x - 6 = 90

3x = 96

x = 32

2x - 6 = 2(32) - 6 = 58

Answer: 32°, 58°

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Answer:

( x + 2 ) and ( x - 11 )

Step-by-step explanation:

x² - 9x - 22 = 0

x² + 2x - 11x - 22 = 0

x ( x + 2 ) - 11 ( x + 2 ) = 0

( x + 2 ) ( x - 11 ) = 0

Factors of x² - 9x - 22 are ( x + 2 ) and ( x - 11 ).

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Andrei has a job in the circus walking on stilts. Andrei is 11/10 meters tall. The foot supports of his stilts are 23/10 meters
vovangra [49]
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You can then simplify by finding the greatest common factor (2)
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3 years ago
Write an equation in slope intercept form for: m = - 1 and b = 2
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y=-1x+2

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g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the
blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean amount of money spent on textbooks = $121.60

            \sigma = population standard deviation = $6.36

            n = sample of students = 36

            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                      of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

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                                      = [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0
3 years ago
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