Answer:
(-162)/7 or -23 1/7 as mixed fraction
Step-by-step explanation:
Simplify the following:
(-36)/14 (-18) (-3)/6
Hint: | Express (-36)/14 (-18) (-3)/6 as a single fraction.
(-36)/14 (-18) (-3)/6 = (-36 (-18) (-3))/(14×6):
(-36 (-18) (-3))/(14×6)
Hint: | In (-36 (-18) (-3))/(14×6), divide -18 in the numerator by 6 in the denominator.
(-18)/6 = (6 (-3))/6 = -3:
(-36-3 (-3))/14
Hint: | In (-36 (-3) (-3))/14, the numbers -36 in the numerator and 14 in the denominator have gcd greater than one.
The gcd of -36 and 14 is 2, so (-36 (-3) (-3))/14 = ((2 (-18)) (-3) (-3))/(2×7) = 2/2×(-18 (-3) (-3))/7 = (-18 (-3) (-3))/7:
(-18 (-3) (-3))/7
Hint: | Multiply -18 and -3 together.
-18 (-3) = 54:
(54 (-3))/7
Hint: | Multiply 54 and -3 together.
54 (-3) = -162:
Answer: (-162)/7
Answer:
34 is the angle in the top left 87 is the angle on the top right ans 59 is the angle on the bottom right
Step-by-step explanation:
Answer:
I think -7; +1
(sorry if I write this but the answer must be at least 20 characters long)
Answer:
There is sufficient evidence that fuel economy goal has been attained.
Step-by-step explanation:
The hypothesis :
H0 : μ < 30.2
H1 : μ ≥ 30.2
The test statistic :
(xbar - μ) ÷ (s/√(n))
xbar = 32.12 ; s = 4.83 ; n = 50
Test statistic :
(32.12 - 30.2) ÷ (4.83/√(50))
1.92 ÷ 0.6830651
T = 2.811
Using the Pvalue from test statistic calculator :
Since we used the sample standard deviation, we use the T distribution
df = n - 1 = 50 - 1 = 49
Pvalue(2.811, 49) ; one tailed = 0.00354
At α = 0.05
Pvalue < α ; then we reject the null and conclude that there is sufficient evidence that fuel economy goal has been attained
