Order each set of values from least to greatest

1. One week Carlos bought 3 bags of Tabitha Tidbits and 4 bags of Figaro Flakes for $43.00. The next week he bought 3 bags of Ta

d1i1m1o1n [39]

Tabitha Tidbits costs $7 per bag, and Figaro Flakes is $5.50 per bag.
You need to set up a system of equations. Use "x" for Tabitha Tidbits and "y" for Figaro Flakes, and let the total cost of each trip equal c. Using the equation ax+by=c, substitute the cost of each trip in for c, and the number of bags for each food for a and b respectively. The two equations will be:
3x+4y=43
3x+6y=54
Isolate x in the first equation and you will get:
x=(43-4y)/3
Substitute the above equation for x into the other equation:
3*((43-4y)/3)+6y=54
Isolate y in this equation, and you will get 11/2, which is 5.5
So the cost of one bag of Figaro Flakes is $5.50
Now substitute this into the equation where you isolated x:
(43-4(5.5))/3
You will get x=7, so a bag of Tabitha Tidbits is $7

3 0

3 years ago

Write the fraction or mixed number as a percent.<br> 1 21/50 =

Eddi Din [679]

242% should be the answer

4 0

3 years ago

Read 2 more answers

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)

Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$