D=√(-9-0)^2+(40-0)^2
d=√1681
d=41
pretty much about the same as before.
a = weight of a large box
b = weight of a small box.
we know their combined weight is 65 lbs, thus a + b = 65.
we also know that the truck has 60 large ones, and 55 small ones, thus 60*a is the total weight for the large ones and 55*b is the total weight for the small ones, and we know that is a total of 3775, 60a + 55b = 3775.
●) -2 (3a - 4b)-6a + 3b
-2 (9a-7b)
●) 10 - 6a - 4y - (-2a) + 9y
10 - 6a - 13y - (-2a)
●) 9 (2x-6) - 3 (5-4x
12 (6x-11)
All I did was combine the numbers with the same variable
so the missing number in the sequence is 34 and 62
Complete question :
A data set includes data from student evaluations of courses. The summary statistics are nequals92, x overbarequals4.09, sequals0.55. Use a 0.10 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Answer:
H0 : μ = 4.25
H1 : μ < 4.25
T = - 2.79
Pvalue =0.0026354
we conclude that there is enough evidence to conclude that population mean is different from 4.25 at 10%
Step-by-step explanation:
Given :
n = 92, xbar = 4.09, s = 0.55 ; μ = 4.25
H0 : μ = 4.25
H1 : μ < 4.25
The test statistic :
T = (xbar - μ) ÷ s / √n
T = (4.09 - 4.25) ÷ 0.55/√92
T = - 0.16 / 0.0573414
T = - 2.79
The Pvalue can be obtained from the test statistic, using the Pvalue calculator
Pvalue : (Z < - 2.79) = 0.0026354
Pvalue < α ; Hence, we reject the Null
Thus, we conclude that there is enough evidence to conclude that population mean is different from 4.25 at 10%