Let x be the number of $10 bills in Iago pocket.
1. If he has twice as many $1 bills as $10 bills, then he has (2x) $1 bills.
2. He has two fewer $20 bills than he does $10 bills, then he has (x-2) $20 bills.
3. He has three more $5 bills than $10 bills, then he has (x+3) $5 bills.
In total he has $160 that is x·10+2x·1+(x-2)·20+(x+3)·5.
Equate these two expressions and solve the equation:
Thus, he has
- 5 bills for $10;
- 10 bills for $1;
- 3 bills for $20;
- 8 bills for $5.
Answer:
Step-by-step explanation:
We can prove that a tangent will always be perpendicular to the radius touching it. So, the other angle in the diagram is 
.
Because all the angles of a triangle sum to 
, we have that 
.
We combine like terms on the left side to get 
.
We subtract 
on both sides to get 
.
So, 
and we're done!
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
Y=3x-5 and passes through the point 5,10 should be 56
