Answer:
The maximum number of pounds of potato salad that Charlie can buy is 0.375
Step-by-step explanation:
see the attached figure to better understand the problem
Let
a ----> the cost of one tuna sandwich
b ----> the cost of a bottle of apple juice
c ----> the cost per pound of potato salad
x ----> pounds of potato salad
we have
we know that
He wants to buy a tuna sandwich, a bottle of apple juice, and x pounds of potato salad and can spend up to $8
The inequality that represent this situation is
substitute the given values
Solve for x
Combine like terms
Subtract 6.50 both sides
Divide by 4 both sides
therefore
The maximum number of pounds of potato salad that Charlie can buy is 0.375
we know that
The multiplicative rate of change of the exponential function between two points is equal to
![[f(b) / f(a) ] / (b-a)](https://tex.z-dn.net/?f=%5Bf%28b%29%20%2F%20f%28a%29%20%5D%20%2F%20%28b-a%29)
Let
we have that
substitute in the formula
![[4 / 6 ] / (2-1)=4/6=2/3](https://tex.z-dn.net/?f=%5B4%20%2F%206%20%5D%20%2F%20%282-1%29%3D4%2F6%3D2%2F3)
Let
we have that
substitute in the formula
![[(16/9) / (8/3) ] / (4-3)=48/72=2/3](https://tex.z-dn.net/?f=%5B%2816%2F9%29%20%2F%20%288%2F3%29%20%5D%20%2F%20%284-3%29%3D48%2F72%3D2%2F3)
therefore
<u>the answer is the option</u>
B.) 2/3
Answer:
Step-by-step explanation:
Kiesha will have more than 97% of the products working.
Kiesha’s experimental probability is 1/50
When the inventory is 4000 clocks, the prediction is that 3920 clocks will work.
Answer:
-5/6
Step-by-step explanation:
-1 2/3=-5/3
-5/3+5/6=-10/6+5/6=-5/6
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.
