Answer:
2 hours
Step-by-step explanation:
30 miles/x hours = 15 miles/1 hour
30 = 15x
x = 2 hours
*Given
Money of Phoebe - 3 times as much as Andy
Money of Andy - 2 times as much as Polly
Total money of Phoebe, - <span>£270
</span> Andy and Polly
*Solution
Let
B - Phoebe's money
A - Andy's money
L - Polly's money
1. The money of the Phoebe, Andy, and Polly, when added together would total <span>£270. Thus,
</span>
B + A + L = <span>£270 (EQUATION 1)
2. Phoebe has three times as much money as Andy and this is expressed as
B = 3A
3. Andy has twice as much money as Polly and this is expressed as
A = 2L</span> (EQUATION 2)
<span>
4. This means that Phoebe has ____ as much money as Polly,
B = 3A
B = 3 x (2L)
B = 6L </span>(EQUATION 3)<span>
This step allows us to eliminate the variables B and A in EQUATION 1 by expressing the equation in terms of Polly's money only.
5. Substituting B with 6L, and A with 2L, EQUATION 1 becomes,
6L + 2L + L = </span><span>£270
</span> 9L = <span>£270
</span> L = <span>£30
So, Polly has </span><span>£30.
6. Substituting L into EQUATIONS 2 and 3 would give us the values for Andy's money and Phoebe's money, respectively.
</span>
A = 2L
A = 2(£30)
A = £60
Andy has £60
B = 6L
B = 6(£30)
B = £180
Phoebe has £180
Therefore, Polly's money is £30, Andy's is £60, and Phoebe's is £180.
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
Making a list can help you solve a problem by:
•making more organized
•it can make it easier to organize data
•helps you stay on track
those are some of the basic ways,
