Answer: the value for the associated test statistic is 1.2653
Step-by-step explanation:
Given that;
sample size one n₁ = 10
mean one x"₁ = 6.4
standard deviation one S₁ = 1.1
sample size two n₁ = 11
mean two x"₂ = 5.6
standard deviation one S₁ = 1.7
H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂
Pooled Variance
sp = √( { [(n₁ - 1) × s₁² + (n₂ - 1) × s₂²] / (n₁ + n₂ - 2)} × (1/n₁ + 1/n₂))
we substitute
= √( { [(10 - 1) × (1.1)² + (11 - 1) × (1.7)²] / (10 + 11 - 2)} × (1/10 + 1/11))
= √( { [(9) × 1.21 + (10) × 2.89] / (19) } × (0.1909))
= √({[ 39.79 ] / 19} × (0.1909))
= √( 2.0942 × 0.1909)
= √( 0.39978 )
= 0.63228
Now Test Statistics will be;
t = ( x"₁ - x"₂) / sp
we substitute
t = ( 6.4 - 5.6) / 0.63228
t = 0.8 / 0.63228
t = 1.2653
Therefore the value for the associated test statistic is 1.2653
Answer:
B
Step-by-step explanation:
Distribute 6 to 5/8u and 1 and you get 15/4u + 6. Distribute -6 to -7/4u and -5 and you get 21/2u and 30. The equation will be 15/4u + 6 + 21/2u + 30 and then combine like terms to get 57/4u + 36.
X² + y² = 225
x - 7y = -75
x = 7y - 75
x² + y² = 225
(7y-75)² + y² = 225
(7y - 75)(7y - 75) + y² = 225
49y² - 525y - 525y + 5625 + y² = 225
50y² - 1050y + 5400 = 0
50(y² - 21y + 108) = 0
y² - 21y + 108 ⇒ (y - 12)(y - 9)
x = 7(12) - 75
x = 84 - 75
x = 9 ⇒ (9,12)
x = 7(9) - 75
x = 63 - 75
x = -12 ⇒ (-12,9)
Answer:
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Answer:
Step-by-step explanation:
If there is 1500 cellular phones per 1000 people, then the number of phones per person is:
Now we know that 
