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:
15C is 59F
Step-by-step explanation:
As an equation, F = (1 4/5)C + 32. 1 and 4/5 is the same as 9/5.
Substitute 15 for C.
F = (9/5)C + 32
F = (9/5)(15) + 32 Multiply 15 by the top number in 9/5
F = 135/5 + 32 Convert 32 to a fraction over 5
F = 135/5 + 160/5 Add the numerators under the same denominator
F = 295/5 Divide
F = 59 Degrees in Farenheit
Therefore 15°C is 59°F.
200
If the number is divisible by 5, then the last digit should be either 5 or 0.
First digit is fixed.
Second digit can be 10 ways.
Third digit can be 10 ways.
Fourth digit can be 2 ways.
So total number of digits 1 x 10 x 10 x 2 = 200
41 it’s really simple you just subtract 3 from 44 and it’s 41