Answer:
$8.78 per pair
Step-by-step explanation:
Answer:
a: We can't determine this from the given information
b: 98%
Step-by-step explanation:
For a:
n = 42
The confidence interval has equal time on each side of µ, so we can add the two end points and divide them by 2 to find the middle of the interaval:
7.4 + 8.6 = 16
16/2 = 8
Now subtract 7.4 from 8 to find the distance from the mean to the end of the interval
8 - 7.4 = 0.6
So the sample mean, plus the calculated error was 0.6 minutes.
We don't have a way of calculating the sample mean with the given information. We could only find the sample standard deviation and the variance.
For b:
We have:
E = 0.6
s = 1.606
n = 42
See attached photo for the calculation of this value
The value is 2.421.
Using a sample size of 42, our degrees of freedom are 41. Use the t-distribution chart to see which level of confidence has 2.421 under it.
The level of confidence is: 98%
We need 41 degrees of freedom, but the chart has only 40, then 45. We can see that 40 has 2.423, and the values go down as the degrees of freedom go up, so 41 will correlate to 2.421
Answer:
Step-by-step explanation:
Since the expressions are multiplied, for the product to equal zero, one term or another has to equal 0 because 0 times anything is zero.
(10-x)(3x-9)=0 when
10-x=0, x=10
3x-9=0, x=9/3=3
So there are two solutions when x=3 and x=10
Hello there!
To solve a function by graphing would be to graph it and then look at where it intersects the x-axis. This way is not always accurate due to the fact that one cannot graph it to the exact point all the time, and even if they can or use a graphing calculator, it's difficult to read the graphs too. For those two reasons, solving by graphing throws off the answer most of the time, therefore this method is rarely reliable and it's important to learn other ways to solve.
I really hope this helps!
Best wishes :)
No. Mass is the measurement of how much matter is in something.
Take neutron stars for example; smaller than our sun but has more mass. Because of density.