Answer:
<em>Total cost=$73.60
</em>
Step-by-step explanation:
<u>Arithmetics To Solve Simple Problems
</u>
Sometimes the use of the most basic operations can be handy to solve problems of a variety of complexities without the use of equations or algebra. We'll approach to the solution of this question by pure logic and basic operations.
If the gift was originally meant to be paid for all 5 brothers, each one would have paid one-fifth of the cost. But since the younger brother won't be paying anything, the cost will be divided among the other four brothers, each one of which will pay now one-fourth of the cost. But this new division is $3.68 more than the previous one. If we take the difference between both fractions, we get
That fraction is the portion of the money they are paying in excess, so the total cost should be twenty times that excess, i.e.
Total cost= 20*3.68=$73.60
Let's check. Each brother should have paid $73.60/5=$14,72. But the four brothers really paid $73.6/4=$18.40. The difference is cost is $18.40-$14,72=$3.68 as required
To find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function.
At the start, you have 10 marbles in total and 6 red marbles. The probability of the first event is 6/10.
Since you do not place the marble back in, you now have 9 marbles in total and 3 blue marbles. This probability of the second event is 3/9.
To find the probability of both events occurring, we multiply the two probabilities together. (6•3)/(10•9)=18/90=1/5
Thus, there is a 20% chance of drawing a red marble then a blue marble without placing the red marble back in the bag.
Using the t-distribution, as we have the standard deviation for the sample, we could claim that the mean weight in pounds is in the interval (79.1, 85.7).
<h3>What is a t-distribution confidence interval?</h3>
The confidence interval is:
In which:
- is the sample mean.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed <em>99% confidence interval</em>, with 100 - 1 = <em>99 df</em>, is t = 2.6259.
The other parameters have values given by:
.
We could claim that the mean is any value in the 99% confidence interval, which has bounds given by:
More can be learned about the t-distribution at brainly.com/question/16162795
T>L+D/B
(Assume no parentheses missing and use PEMDAS rule)
Subtract L from both sides
T-L > L-L + D/B
T-L>D/B
Multiply both sides by B (assuming B>0)
B(T-L)>D/B*B
B(T-L)>D
Interchange two sides,
D<B(T-L) (if B>0)
if B<0, then D>B(T-L)