The patient is recommended to take 24 ounces of coffee per day.
<u>Step-by-step explanation:</u>
Patient normal intake of coffee daily = 5 times 12 ounces
= 5 × 12 = 60 ounces
Doctor recommends to cut off 60 % of the normal intake of coffee.
And advised to take only 40 % of the daily intake of coffee.
So 60 % of 60 ounces can be calculated as,
= 36 ounces
40 % of 60 ounces is 24 ounces.
So 60 - 36 = 24 ounces of coffee per day
Answer:
in this method a variable is expressed in terms of another variable from one equation and it is substituted in remaining equation.
Step-by-step explanation:
solve:x+2y=9 and 3x-y=13.
x+ 2y=9…be the first equation
3x-y=13…be the second equation
from first equation
x=9-2y…be the third equation substituting value of x from third equation in second equation, we get
3(9-2y)-y=13
or, 27-6y-y=13
or, y=2
now,
substituting value of y in third equation, we get,
x=9-2×2
=9-4
=5
the required value of x and y are 5&2
Step-by-step explanation:
z^7/z^-14=21
z^21=21
Subtract 8p from 9p and then rewrite the equation. Then you add 7 to 4 and rewrite the equation again. Then you divide 1p by 11 and get p=11.
Answer:
Step-by-step explanation:
Hello!
The chemistry instructor tested the hypothesis that the proportion of students that passed the introductory chemistry class is better with an embedded. If the known proportion for this population is 65%, the tested hypothesis is:
H₀: p=0.65
H₁: p>0.65
The calculated statistic is Z=2.52 and the associated p-value: 0.0059
Remember:
The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
In this case:
P(Z≥2.52)=0.0059
There is no significance level, the most common one is α: 0.05 so I'll use it as an example.
To make a decision using the p-value you have to compare it to the α.
If p- value>α then you support the null hypothesis (In this case, you can say that there is no change in the proportion of students that passed the introductory chemistry class with an embedded tutor.)
If p-value≤α your decision will be to reject the null hypothesis (In this case, there is significant evidence to say that there is an improvement in the success rate of the introductory chemistry class with an embedded tutor?
Since the p-value:0.0059 is less than the significance level 0.05, you will decide to reject the null hypothesis.
I hope you have a SUPER day!
