Ask question

Ask question

All categories

3 years ago

14

The length of the sides of a triangular garden are 10m 11m 13m.Find the measure of the angle opposite the longest side to the ne

arest tenth of a degree

1 answer:

Kaylis [27]3 years ago

3 0

C^2=a^2+b^2-2abcosC
13^2=10^2+11^2-2×10×11×cos(°)
169=100+121-220cos(°)
220cos(°)=100+121-169
220cos(°)=52
cos(°)= 13/55
°=cos-1(13/55)
°=76.328°

You might be interested in

The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles.

Free_Kalibri [48]

Answer:

1191.4 ; 34.5

Step-by-step explanation:

Given the data:

29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150

The sample variance and standard deviation can be obtained thus :

Σ(X - m)² / (n - 1)

Where, m = mean of the sample

n = sample size

The standard deviation equals, sqrt(variance )

Using a calculator:

The variance, σ² ;

Mean = Σx / n = 1681 / 20 = 84.05

(x -m)^2

[(29-84.05)^2 + (37-84.05)^2 + (38-84.05)^2 + (40-84.05)^2 + (58-84.05)^2 + (67-84.05)^2 + (68-84.05)^2 + (69-84.05)^2 + (76-84.05)^2 + (86-84.05)^2 + (87-84.05)^2 + (95-84.05)^2 + (96-84.05)^2 + (96-84.05)^2 + (99-84.05)^2 + (106-84.05)^2 + (112-84.05)^2 + (127-84.05)^2 + (145-84.05)^2 + (150-84.05)^2] / 19

22636.95 / 19

= 1191.4184 = 1191.42

Standard deviation = sqrt( Variance)

Standard deviation = sqrt(1191.4184)

Standard deviation = 34.516929 = 34.52

4 0

3 years ago

PLEASE HELP!! Sue and Cat have $20 for an Uber ride home. It costs $3.00 per mile plus a $2 fixed charge. What is the maximum nu

m_a_m_a [10]

Answer:

6miles

Step-by-step explanation

5 0

3 years ago

A tobacco company claims that the amount of nicotine in its cigarettes is a randomvariable with mean 2.2 mg and standard deviati

Genrish500 [490]

Answer:

0.0000

Unusual

Step-by-step explanation:

Given that a tobacco company claims that the amount of nicotine in its cigarettes is a random variable with mean 2.2 mg and standard deviation .3 mg.

i.e. population parameters are

$\mu =2.2 \\s = 0.3$

The approximate probability that the sample meanwould have been as high or higher than 3.1

$=P(X\geq 3.1)\\=P(Z\geq \frac{3.1-2.2}{\frac{0.3}{\sqrt{100} } } )\\=P(Z\geq 30)\\$

=0.0000

8 0

3 years ago

The product of 7 and a number, all subtracted from 10

expeople1 [14]

Answer: i think the answer is -3

Step-by-step explanation:

7 0

3 years ago

Helpppppppp me pleaseeeeeeee

andreev551 [17]

Answer:

I think its the last one thats not showing if I'm wrong in sorry

3 0

3 years ago