Suppose the spinner lands on <em>a</em>. There's a 1/3 chance that it'll land on <em>a</em> the second time.
Suppose the spinner lands on <em>b</em>. There's a 1/3 chance that it'll land on <em>b</em> the second time.
Suppose the spinner lands on <em>c</em>. There's a 1/3 chance that it'll land on <em>c</em> the second time.
We've covered all possibilities for the first spin, and they're all equal, so their average is 1/3.
The probability that it'll land on the same letter twice is 33.3%.
At the end of tomorrow, inventory of 5+20 pallets will have been used to fill orders for 2+10 pallets. That leaves inventory on hand of
... 25 - 12 = 13 pallets
Since each holds 20 panels, that amounts to
... (13 pallets) × (20 panels/pallet) = 260 panels . . . . remaining
Answer:
z= 0.278
Step-by-step explanation:
Given data
n1= 60 ; n2 = 100
mean 1= x1`= 10.4; mean 2= x2`= 9.7
standard deviation 1= s1= 2.7 pounds ; standard deviation 2= s2 = 1.9 lb
We formulate our null and alternate hypothesis as
H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)
We set level of significance α= 0.05
the test statistic to be used under H0 is
z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂
the critical region is z > ± 1.96
Computations
z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100
z= 10.4- 9.7/ √ 7.29/60 + 3.61/100
z= 0.7/√ 0.1215+ 0.0361
z=0.7 /√0.1576
z= 0.7 (0.396988)
z= 0.2778= 0.278
Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0 at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.
Answer:
A. f(s) = 2s + 34
B. Mole
C. f(s) = 2s + 54
Step-by-step explanation:
A. Its a prediction the 34 can be any number reasonably close to 25
B. Amount of substance - mole (mole)
c. f(s) = 2s + 54
