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Natali [406]
3 years ago
10

PLEASE HELP ME I NEED IT RN​

Mathematics
1 answer:
lapo4ka [179]3 years ago
5 0

Answer:

(47)  c. given

(48) b. definition of complementary

(49)  a, m<2 = m<3

(50) d. definition of congruent angles

Step-by-step explanation:

From the given question, the proof can be completed as;

Statement                                          Reason

1. <2 and < 1 are complementary     (47)  c. given

   <3 and <1 are complementary

2. m<2 + m<1 = 90^{o}                             (48) b. definition of complementary angles

   m<3 + m<1 = 90^{o}

3. m<2 + m<1 = m<3 + m<1                 3. Substitution property

4. (49)  a, m<2 = m<3                         4. Subtraction property

5. <2 ≅ <3                                           (50) d. definition of congruent angles

Therefore,

(47)  c. given

(48) b. definition of complementary

(49)  a, m<2 = m<3

(50) d. definition of congruent angles

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A researcher wishes to estimate the average blood alcohol concentration​ (BAC) for drivers involved in fatal accidents who are f
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Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 ​g/dL.

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

                               P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~   t_n_-_1

where, \bar X = sample mean BAC = 0.16 g/dL

            s = sample standard deviation = 0.080 ​g/dL

            n = sample of drivers = 60

            \mu = population mean BAC in fatal crashes

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

So, a 90% confidence interval for the population mean, \mu is;

P(-1.672 < t_5_9 < 1.672) = 0.90  {As the critical value of t at 59 degrees of

                                              freedom are -1.672 & 1.672 with P = 5%}    P(-1.672 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.672) = 0.90

P( -1.672 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.672 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.672 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.672 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.672 \times {\frac{s}{\sqrt{n} } } , \bar X+1.672 \times {\frac{s}{\sqrt{n} } } ]

                                       = [ 0.16-1.672 \times {\frac{0.08}{\sqrt{60} } } , 0.16+1.672 \times {\frac{0.08}{\sqrt{60} } } ]

                                       = [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

7 0
4 years ago
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