How do you find the number of possible choices for a 2 digit password that is greater than 19?

1 answer:

3 0

There probably is a formula to do this, but a simple way to find this one is to just think of all two digit numbers. That is 00-99. All you do is count the number of possible choices from 20-99. So there are 80 possible choices.

You might be interested in

a pair of socks costs $5.00. a package of 3 pairs costs $12.00. how much would you save if you buy 2 packages instead of 6 indiv

svp [43]

you will save 6 dollars If you buy 2 packages, instead if 6 individual pairs.

7 0

4 years ago

beth won 45 lollipops playing the bean bag toss after giving some away she had only 28 remaining how many did she give away

Finger [1]

The answer is 17 45-28=17

7 0

3 years ago

On a coordinate plane, 2 exponential functions are shown. f (x) decreases from quadrant 2 to quadrant 1 and approaches y = 0. It

aliina [53]

Answer: $-4\left(\frac{1}{2} \right)^{x}$

Step-by-step explanation:

To reflect the graph of a function across the x-axis,

$y=f(x) \longrightarrow y=-f(x)$

3 0

2 years ago

-2.9=(x-1200)/ 21.91

nirvana33 [79]

Answer:

Its A i just took the test

Step-by-step explanation:

5 0

3 years ago

Each variable indicates different weights. Which weight can you find? Find it.

SOVA2 [1]

Answer:

We can only be certain that a weighs 12.

There are infinitely many possiblities for b and c.

Step-by-step explanation:

We have the equation:

$a+b+c+a+c+b+a+c=12+a+a+b+b+c+c+c$

Each variable indicates a weight.

We would like to determine the weights of each variable (if possible).

First, we can rearrange the equation to acquire:

$(a+a+a)+(b+b)+(c+c+c)=12+(a+a)+(b+b)+(c+c+c)$

We can combine like terms:

$3a+2b+3c=12+2a+2b+3c$

Notice that both sides have 2b and 3c. Therefore, it is possible for us to cancel them since each nullify the other side. So, we will subtract 2b and 3c from both sides. This yields:

$3a=12+2a$