There really isn’t a possible answer since there is not enough information given. Not from what I have learned so far at least, but using common sense, and the process of elimination I would say 30
6.3 many more cups of dry food will Maria's pet have eaten than Trenton's pet will have eaten over 2 seven-day weeks
<u>Step-by-step explanation:</u>
We have , Trenton and Maria record how much dry food their pets eat on average each day.• Trenton's pet: 4/5 cup of dry food• Maria's pet: 1.25 cups of dry food. Based on these averages . We need to find how many more cups of dry food will Maria's pet have eaten than Trenton's pet will have eaten over 2 seven-day weeks . We need to find how much they eat for 14 days as:
Trenton's pet: 4/5 cup of dry food•
With 4/5 per day , for 14 days :
⇒ 
⇒ 
⇒ 
Maria's pet: 1.25 cups of dry food.
With 1.25 per day , for 14 days :
⇒ 
⇒ 
Subtracting Maria's - Trenton's :
⇒ 
That means , 6.3 many more cups of dry food will Maria's pet have eaten than Trenton's pet will have eaten over 2 seven-day weeks
Let x = speed and y = distance
y = x * 45
y = (x - 4) * 70
45x = 70x - 280
-25x = -280
x = 11.2
Put it back into the first equation:
y = 11.2 x 45
y = 504miles
Hope this helps! Any questions let me know :)
step-by-step explanation: the increase = 204 - 120
increase = 84
percent increase = 84/120
percent increase = 70% so of course the percentage is 70% ( I hope this helps you )
Answer:
Type I: 1.9%, Type II: 1.6%
Step-by-step explanation:
given null hypothesis
H0=the individual has not taken steroids.
type 1 error-falsely rejecting the null hypothesis
⇒ actually the null hypothesis is true⇒the individual has not taken steroids.
but we rejected it ⇒our prediction is the individual has taken steroids.
typr II error- not rejecting null hypothesis when it has to be rejected
⇒actually null hypothesis is false ⇒the individual has taken steroids.
but we didnt reject⇒the individual has not taken steroids.
let us denote
the individual has taken steroids by 1
the individual has not taken steroids.by 0
predicted
1 0
actual 1 98.4% 1.6%
0 1.9% 98.1%
so for type 1 error
actual-0
predicted-1
therefore from above table we can see that probability of Type I error is 1.9%=0.019
so for type II error
actual-1
predicted-0
therefore from above table we can see that probability of Type I error is 1.6%=0.016
