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:It is -6, 0
Step-by-step explanation:
Distance between A and B
= Distance between C and D
= sqrt((4 - 1)^2 + (5 - 2)^2)
= sqrt(3^2 + 3^2)
= sqrt(2 * 3^2)
= sqrt(3^2) * sqrt(2)
= 3sqrt(2)
Distance between B and C
= Distance between A and D
= sqrt((4 - 3)^2 + (5 - 0)^2)
= sqrt(1^2 + 5^2)
= sqrt(26)
Since sqrt(26) is more than 3sqrt(2), the length must be sqrt(26).
Hope this helps you.
The rate of the wind is given to be 210 the rate of the plane in mid air is given as 830
<h3>Howe to solve for the rate of the plane and wind</h3>
In 7 hours the kilometer traveled = 4340
P + W = 4340/7
= 620
P-W = 3120/3 hours
= 1040
2p = 1040 + 620
= 1660
p =830
W = 1040 - 830
= 210
Hence the rate of the wind is given to be 210 the rate of the plane in mid air is given as 830
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brainly.com/question/24304697
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