<span><span><span>2x</span>−1</span>≤<span>x5
</span></span>Let's find the critical points of the inequality.
<span><span><span>2x</span>−1</span>=<span>x^5
</span></span><span><span><span><span>2x</span>−1</span>−<span>x^5</span></span>=<span><span>x^5</span>−<span>x^5 </span></span></span>(Subtract x^5 from both sides)
<span><span><span><span>−<span>x^5</span></span>+<span>2x</span></span>−1</span>=0
</span><span><span><span>(<span><span>−x</span>+1</span>)</span><span>(<span><span><span><span><span>x^4</span>+<span>x^3</span></span>+<span>x^2</span></span>+x</span>−1</span>)</span></span>=0 </span>(Factor left side of equation)
<span><span><span><span>−x</span>+1</span>=<span><span><span><span><span><span>0<span> or </span></span><span>x^4</span></span>+<span>x^3</span></span>+<span>x^2</span></span>+x</span>−1</span></span>=0 </span>(Set factors equal to 0)
<span><span><span>x=<span><span>1<span> or </span></span>x</span></span>=<span><span>0.51879<span> or </span></span>x</span></span>=<span>−<span>1.290649</span></span></span>
Answer:
Cool
Do you need help with anything or is it just a fact?
For a binomial experiment in which success is defined to be a particular quality or attribute that interests us, with n=36 and p as 0.23, we can approximate p hat by a normal distribution.
Since n=36 , p=0.23 , thus q= 1-p = 1-0.23=0.77
therefore,
n*p= 36*0.23 =8.28>5
n*q = 36*0.77=27.22>5
and therefore, p hat can be approximated by a normal random variable, because n*p>5 and n*q>5.
The question is incomplete, a possible complete question is:
Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
Suppose n = 36 and p = 0.23. Can we approximate p hat by a normal distribution? Why? (Use 2 decimal places.)
n*p = ?
n*q = ?
Learn to know more about binomial experiments at
brainly.com/question/1580153
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Answer:
We use letters to represent a variable in expressions. I hope this helps!
Step-by-step explanation:
