1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok
2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No
3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No
4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok
5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No
Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10
Answer:
45%
Step-by-step explanation:
45 boxes are shaded
so percentage shaded is 45 out of the total which is 100
so 45 / 100 X 100
=45%
Answer:
32, because 8*4 is 32
Step-by-step explanation:
Answer:
172-602 I think
Step-by-step explanation:
Answer:
0.818
Step-by-step explanation:
Since the shipment has a ton of aspirin tablets, we can assume that we pick 13 of them <em>with</em> <em>reposition, </em>because the probability shoudn't change dramatically from the probability of picking without reposition if we do so.
We call D the amount of defective tablets. If we assume that we pick the tablets with reposition, then we obtain that D is a random variable of Binomial distribution with parameters 13 and 0.6 (the probability of picking a defective tablet).
We want D to be at most one. To calculate the probability of that event we add up the probability of D being equal to 0 and the probability of D being equal to one. Since D is binomial, we have
We conclude that

Hence, the shipment will be accepted with probability 0.818
<em>I hope this helps you!</em>