Answer:
one glass of milk = 15 carbs
one bar = 14 carbs
Step-by-step explanation:
let 'm' = one glass of milk
let 'b' = one snack bar
system of equations:
2m + 3b = 72
4m + 2b = 88
I used the elimination method and multiplied the 1st equation by -2
-4m - 6b = -144
+ <u>4m + 2b = 88</u>
-4b = -56
b = 14
now solve for 'm'
2m + 3(14) = 72
2m + 42 = 72
2m = 30
m = 15
73 minus 24 plus 19 gives you 68 people left on the train.
4x² + 2x - 30 = 0
<span>
factor out the GCF:
</span>2(2x² + x - 15) = 0
<span>
factor the trinomial completely:
2x</span>² + x - 15 = 0
2x² + 6x - 5x - 15 = 0
2x(x + 3) - 5(x + 3) = 0
(2x - 5)(x + 3) = 0
<span>use the zero product property and set each factor equal to zero and solve:
2x - 5 = 0 or x + 3 = 0
2x = 5 x = -3
x = 2.5
</span><span>The roots of the function are x=-3, x=2.5</span>
Answer:
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Step-by-step explanation:
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
Where u1 is the group tested after they were awake for 24 hours.
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.05, 13) = 1.77
Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13
Here the difference between the sample means is x`1- x`2= 35-24= 11
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/13 [ 120²+360²]
Sp = 105.25
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= 11/ 105.25 √1/5+ 1/10
t= 11/57.65
t= 0.1908
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
