Answer:
the length is 4
Step-by-step explanation:
12-8 =4
hope it helps
Our "y-intercept" (where x = 0) acts as the foundation of our function here.
We can extrapolate from what we have here...if x is 0, y is going to be 2.
So, we start at 2 for y, and every time we decrease x by 2, y increases by 1.
If we decrease x by just 1, y increases by ½.
This also goes in the opposite direction. Increase x by 1, decrease y by ½.
We can write the change in y as -½x.
That change is being added to our original value of 4, of course.
The equation of this line is
.
<span><span>0.1n</span>=89.5</span>
<span /><span>Step 1: Divide both sides by 0.1.</span>
<span><u>Answer: </u><span><span>n=89</span></span></span>
<span><span><span /><span><span><span><span>(10)</span><span>(3)</span></span>n</span>=630</span></span></span>
<span><span><span /><span><span>Step 1: Simplify both sides of the equation.</span></span></span></span>
<span><span><span /></span></span>30n= 630
step1: divide both sides by 30.
<u>Answer</u>:<u /> n= 21
$78
The difference between choice A and B is 5 meals and $80 so 1 meal is $16. So we can subtract $16 from $250 and we get $234 for 3 nights. 1 night is $78
Answer:
<em>H₀</em>: <em>μ</em>₁ = <em>μ</em>₂ vs, <em>Hₐ</em>: <em>μ</em>₁ > <em>μ</em>₂.
Step-by-step explanation:
A two-sample <em>z</em>-test can be performed to determine whether the claim made by the owner of pier 1 is correct or not.
It is provided that the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations.
The hypothesis to test whether the average weights of the fish in pier 1 is more than pier 2 is as follows:
<em>H₀</em>: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. <em>μ</em>₁ = <em>μ</em>₂.
<em>Hₐ</em>: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. <em>μ</em>₁ > <em>μ</em>₂.
The significance level of the test is:
<em>α</em> = 0.05.
The test is defined as:
The decision rule for the test is:
If the <em>p</em>-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.