Answer:
849,400
Step-by-step explanation:
hi
According to BODMAS we solvethe equation:-
B- brackets
O- of
D-divide
M- multiple
A-Add
S- subtract
=> 23 - 4 - 6 + 7 + 5 - 1 x 0 + 1 = x
=> x=26
HOPE IT HELPS YOU.
Answer:
The answer is 4.
Step-by-step explanation:
<h2><u><em>
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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:
(I suppose that we want to find the probability of first randomly drawing a red checker and after that randomly drawing a black checker)
We know that we have:
12 red checkers
12 black checkers.
A total of 24 checkers.
All of them are in a bag, and all of them have the same probability of being drawn.
Then the probability of randomly drawing a red checkers is equal to the quotient between the number of red checkers (12) and the total number of checkers (24)
p = 12/24 = 1/2
And the probability of now drawing a black checkers is calculated in the same way, as the quotient between the number of black checkers (12) and the total number of checkers (23 this time, because we have already drawn one)
q = 12/23
The joint probability is equal to the product between the two individual probabilities:
P = p*q = (1/2)*(12/23) = 0.261
T
