Answer:
<u><em>Dear its two variable function so either you should give me other function of x and y or you should give me condition to solve this problem</em></u>
<u><em>so see you next time</em></u>
Step-by-step explanation:
As soon as I read this, the words "law of cosines" popped
into my head. I don't have a good intuitive feeling for the
law of cosines, but I went and looked it up (you probably
could have done that), and I found that it's exactly what
you need for this problem.
The "law of cosines" relates the lengths of the sides of any
triangle to the cosine of one of its angles ... just what we need,
since we know all the sides, and we want to find one of the angles.
To find angle-B, the law of cosines says
b² = a² + c² - 2 a c cosine(B)
B = angle-B
b = the side opposite angle-B = 1.4
a, c = the other 2 sides = 1 and 1.9
(1.4)² = (1)² + (1.9)² - (2 x 1 x 1.9) cos(B)
1.96 = (1) + (3.61) - (3.8) cos(B)
Add 3.8 cos(B) from each side:
1.96 + 3.8 cos(B) = 4.61
Subtract 1.96 from each side:
3.8 cos(B) = 2.65
Divide each side by 3.8 :
cos(B) = 0.69737 (rounded)
Whipping out the
trusty calculator:
B = the angle whose cosine is 0.69737
= 45.784° .
Now, for the first time, I'll take a deep breath, then hold it
while I look back at the question and see whether this is
anywhere near one of the choices ...
By gosh ! Choice 'B' is 45.8° ! yay !
I'll bet that's it !
1 stick of butter :))))))))))))))
Answer:
I would say 340 grams........
Step-by-step explanation:
340 grams
Answer: 0.0548
Step-by-step explanation:
Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.
Let 
represents the sample mean GPA for each student.
Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:
![P(\overline{X}>3.42)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.42-3.5}{\dfrac{0.5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-0.08}{\dfrac{0.5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>1.6)\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28%5Coverline%7BX%7D%3E3.42%29%3DP%28%5Cdfrac%7B%5Coverline%7BX%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%3E%5Cdfrac%7B3.42-3.5%7D%7B%5Cdfrac%7B0.5%7D%7B%5Csqrt%7B100%7D%7D%7D%29%5C%5C%5C%5C%3DP%28Z%3E%5Cdfrac%7B-0.08%7D%7B%5Cdfrac%7B0.5%7D%7B10%7D%7D%29%5C%20%5C%20%5C%20%5BZ%3D%5Cdfrac%7B%5Coverline%7BX%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5D%5C%5C%5C%5C%3DP%28Z%3E1.6%29%5C%5C%5C%5C%3D1-P%28Z%3C1.6%29%5C%5C%5C%5C%3D1-0.9452%3D0.0548)
hence, the required probability is 0.0548.
