One very simple way is to put it in a Ti-84 calculator which shows that cos(105) is -0.2409590492 and sin(15) as 0.6502978402 which you can find online.

8 0

4 years ago

Wingspans of adult horses have approximate normal distribution with mean 125 cm and standard deviation 12 cm. What are the rst a

blagie [28]

Answer:

The first and third quartile of the wingspans are 116.9 cm and 133.1 cm respectively.

Step-by-step explanation:

The first and third quartile of a normal distribution are:

$Q_{1}=\mu-0.675\sigma\\\\Q_{3}=\mu+0.675\sigma$

The information provided is as follows:

$\mu=125\\\sigma=12$

Compute the first and third quartile of the wingspans as follows:

$Q_{1}=\mu-0.675\sigma=125-(0.675\times 12)=116.9\\\\Q_{3}=\mu+0.675\sigma=125+(0.675\times 12)=133.1$

Thus, the first and third quartile of the wingspans are 116.9 cm and 133.1 cm respectively.

7 0

3 years ago