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g100num [7]
3 years ago
5

Given a triangle with a = 12, A = 13°, and B = 12°, what is the length of c? Round to the nearest tenth.

Mathematics
1 answer:
Leni [432]3 years ago
8 0

Answer:

=5.1 units

Step-by-step explanation:

To use the sine rule to find the length of c we first need to find the angle at C.

C=180-(13+12)

C=180-25

C=155°

a/Sin A=c/Sin C

Calculating for the value of c using the sine rule:

12/Sin 13=c/Sin 155

c=12 sin 155

=5.07 units

Thus to 1 decimal places, the value of c is 5.1 units.

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3 years ago
Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:
Vikentia [17]

Answer:

a

   \= x  = 18.5  ,  \sigma =  5.15

b

 15.505 < \mu <  21.495

c

 14.93 < \mu <  22.069

Step-by-step explanation:

From the question we are are told that

    The  sample data is  21, 14, 13, 24, 17, 22, 25, 12

     The sample size is  n  = 8

Generally the ample mean is evaluated as

        \= x  =  \frac{\sum x  }{n}

        \= x  =  \frac{  21 + 14 + 13 + 24 + 17 + 22+ 25 + 12  }{8}

         \= x  = 18.5

Generally the standard deviation is mathematically evaluated as

         \sigma =  \sqrt{\frac{\sum (x- \=x )^2}{n}}

\sigma =  \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}

\sigma =  5.15

considering part b

Given that the confidence level is  90% then the significance level is evaluated as

         \alpha  =  100-90

         \alpha  = 10\%

         \alpha  = 0.10

Next we obtain the critical value of  \frac{ \alpha }{2}  from the normal distribution table the value is  

     Z_{\frac{ \alpha }{2} }  =  1.645

The margin of error is mathematically represented as

      E =  Z_{\frac{ \alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>    E =1.645  *  \frac{5.15 }{\sqrt{8} }

=>     E =  2.995

The 90% confidence interval is evaluated as

       \= x  -  E < \mu <  \= x +  E

substituting values

       18.5 -  2.995 < \mu <  18.5 +  2.995

       15.505 < \mu <  21.495

considering part c

Given that the confidence level is  95% then the significance level is evaluated as

         \alpha  =  100-95

         \alpha  = 5\%

         \alpha  = 0.05

Next we obtain the critical value of  \frac{ \alpha }{2}  from the normal distribution table the value is  

     Z_{\frac{ \alpha }{2} }  =  1.96

The margin of error is mathematically represented as

      E =  Z_{\frac{ \alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>    E =1.96  *  \frac{5.15 }{\sqrt{8} }

=>     E = 3.569

The 95% confidence interval is evaluated as

       \= x  -  E < \mu <  \= x +  E

substituting values

       18.5 - 3.569 < \mu <  18.5 +  3.569

       14.93 < \mu <  22.069

8 0
3 years ago
What is 95% of 120 games?
Basile [38]
The answer to your question is 114
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4 years ago
Read 2 more answers
ANSWER QUICK PLSS!!!
meriva
The correct answer is letter C
4 0
3 years ago
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