The answer is b, hope I helped (:
Answer:
be rewritten as a square of a sum
Step-by-step explanation:
Answer:
Step-by-step explanation:
H0 : μ = 46300
H1 : μ > 46300
α = 0.05
df = n - 1 = 45 - 1 = 44
Critical value for a one tailed t-test(since population standard deviation is not given).
Tcritical = 1.30
The test statistic :(xbar - μ) ÷ (s/sqrt(n))
The test statistic, t= (47800-46300) ÷ (5600√45)
t = 1500
t = 1500 / 834.79871
t = 1.797
The decision region :
Reject H0: if t value > critical value
1. 797 > 1.30
Tvalue > critical value ; We reject H0
Hence, there is sufficient evidence to conclude that cost has increased.
Answer:
1. 5/3 cm
2. 10 in
Step-by-step explanation:
1. The length of a rectangle is 4 cm more than 3 times its width.
Let the length be L and the width B
Then,
L = 4 + 3B
If the area of the rectangle is 15 cm2
Area = LB = 15
B(4 +3B) = 15
3B² + 4B - 15 = 0
3B² - 5B + 9B - 15 = 0
B(3B - 5) + 3(3B - 5) = 0
(3B - 5)(B + 3) = 0
3B - 5 = or B + 3 = 0
B = 5/3 or -3
since B cannot be negative, B = 5/3 cm
2. The ratio of length to width in a rectangle is 2 ∶ 3.
Let the length be L and the width B
Then,
L/B = 2/3
2B = 3L, B = (3/2)L
when the area is 150 in2.
Area = LB = 150
(3/2)L² = 150
L² = 150×2÷3
L² = 100
L = √100
L = ±10
since the length cannot be negative, L = 10 in
D = 2C
J = 1/3D
J = C - 1
J = 1/3D
J = 1/3(2C)
J = 2/3C
2/3C = C - 1
2/3C - 3/3C = -1
-1/3C = -1
C = -1 * -3
C = 3...there are 3 cats
J = C - 1
J = 3 - 1
J = 2.....there are 2 collars
