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2 years ago

In △ABC,c=71, m∠B=123°, and a=65. Find b. A. 101.5 B. 117.8 C. 123.0 D. 119.6

Mathematics

1 answer:

tia_tia [17]2 years ago

8 0

Answer:

Option D

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other the given angle is the included angle. This means that the angle B is formed by the intersection of the lines a and c. Therefore, cosine rule will be used to calculate the length of b. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that c=71, B=123°, and a=65. Plugging in the values:

b^2 = 65^2 + 71^2 - 2(65)(71)*cos(123°).

Simplifying gives:

b^2 = 14293.0182932.

Taking square root on the both sides gives b = 119.6 (rounded to the one decimal place).

This means that the Option D is the correct choice!!!

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Since the two triangles are congruent, which means they are the same shape and size, they share the same angle measures. Since angle B is the same as angle F, it equals 40 degrees.

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6(4x-1) is equivalent to 24x-6 and (2*12x)-(2*3)

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An equivalent expression is an expression which either simplified or factored is equal to it.

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A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st

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Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% 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 credit hours per quarter = 15.2 credit hours

s = sample standard deviation = 2.3 credit hours

n = sample of students = 250

$\mu$ = population mean credit hours per quarter

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

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

P(-1.96 < $t_2_4_9$ < 1.96) = 0.95 {As the critical value of t at 249 degree of

freedom are -1.96 & 1.96 with P = 2.5%}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.2-1.96 \times {\frac{2.3}{\sqrt{250} } }$ , $15.2+1.96 \times {\frac{2.3}{\sqrt{250} } }$ ]

= [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.

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