A diagonal of a parallelogram is 20 inches long and makes angles of 17° and 49° with the sides. How long is the longest side?

1 answer:

3 0

Angles of the parallelogram which is given in the problem are:α = 49° + 17 °= 66°

β = 180° - 66° = 114°

x - length of the longest side of the parallelogram;

Use the Sine Law:x / sin 49° = 20/ sin 17°

x / 0.7547 = 20 / 0.2924

0.2924 x = 15.094

x = 15.094 : .2924

x = 51.62 is the longest side of the parallelogram.

You might be interested in

Mr. Lau drove for 4 hours at a constant speed of r miles per hour.

Ivanshal [37]

Answer:

54 m/h

Step-by-step explanation:

216 ÷ 4

it is simple maths my dude

6 0

3 years ago

x^2+2x+7=21 the approximate value of the greatest solutoon to the equation, rounded to the nearest hundreth

daser333 [38]

9514 1404 393

Answer:

x ≈ 2.87

Step-by-step explanation:

We can subtract 6 to put the equation into a form easily solved.

x^2 +2x +1 = 15

(x +1)^2 = 15

x +1 = √15 . . . . . for the greatest solution, we only need the positive root

x = √15 -1 ≈ 2.87

4 0

3 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$