Considering the situation described, we have that:
a) The critical value is of z = -1.645.
b) Since the test statistic is less than the critical value, we should reject the null hypothesis H0.
<h3>What is the critical value?</h3>
We have a left-tailed test, as we are testing if the proportion is less than a value. Hence the critical value is z with a p-value equals to the significance level, hence z with a p-value of 0.05, hence z = -1.645.
<h3>What is the decision?</h3>
Considering the test statistic, for a left-tailed test, we have that:
- Less than the critical value: Reject H0.
- Equal or greater: Do not reject.
In this problem, z = -2.39 is less than -1.645, hence we should reject the null hypothesis H0.
More can be learned about the test of an hypothesis at brainly.com/question/13873630
#SPJ1
Answer:
width=3
Step-by-step explanation:
area=length X width
length=4 X width
=>area=4 X width X width
=>area=(2 X width)^2
=>36=(2 X width)^2
=>6=2 X width
=>width=3
Yes,
y=x+3
if x=0,
y=0+3
y=3
therefore (0,3) is a solution
We are asked in the problem to evaluate the integral of <span>(cosec^2 x-2005)÷cos^2005 x dx. The function is an example of a complex function with a degree that is greater than one and that uses special rules to integrate the function via the trigonometric functions. For example, we integrate
2005/cos^2005x dx which is equal to 2005 sec^2005 x since sec is the inverse of cos. The integral of this function when n >3 is equal to I=</span><span>∫<span>sec(n−2)</span>xdx+∫tanx<span>sec(n−3)</span>x(secxtanx)dx
Then,
</span><span>∫tanx<span>sec(<span>n−3)</span></span>x(secxtanx)dx=<span><span>tanx<span>sec(<span>n−2)</span></span>x/(</span><span>n−2)</span></span>−<span>1/(<span>n−2)I
we can then integrate the function by substituting n by 3.
On the first term csc^2 2005x / cos^2005 x we can use the trigonometric identity csc^2 x = 1 + cot^2 x to simplify the terms</span></span></span>
