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vivado [14]
3 years ago
7

The identification code on a bank card consists of 1 digit followed by 2 letters. The code must meet the following conditions: T

he digit must be odd. The letters A, E, I, O, and U cannot be used. Letters cannot be used more than once.
Mathematics
1 answer:
jasenka [17]3 years ago
8 0

Answer:

For the code we have 3 selections.

The first selection is a digit that must be odd, so the options are {1, 3, 5, 7 ,9}

So we have 5 options.

The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)

So here we have 27 - 5 = 22 options

The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.

Here we have 22 - 1 = 21 options.

The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:

C = 5*22*21 = 2310.

There are 2310 different possible codes

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4 0
3 years ago
Read 2 more answers
I really need help finding the solution!!
Mrac [35]

Answer:

-0.5 is not a solution

2 is a solution

Step-by-step explanation:

To check if something is a solution of the equation, you need to substitute the value in place of the variable

15 + 2y = -12 - 4y

15 + 2(-0.5) = -12 - 4(-0.5)

15 - 1 = -12 + 2

14 ≠ -10 ∴ -0.5 is not a solution

5 - 2(3x + 5) = 3 - 10x

5 - 2[3(2) + 5] = 3 - 10(2)

5 - 2(6 + 5) = 3 - 10(2)

5 - 12 - 10 = 3 - 120

-17 = -17 ∴ 2 is a solution

3 0
3 years ago
I need to know how to find the slope of the line for #2
son4ous [18]
Slope formula : (y2 - y1) / (x2 - x1)
now pick 2 points on ur line...
(1,2)......x1 = 1 and y1 = 2
(3,0)......x2 = 3 and y2 = 0
now sub
slope = (0 - 2) / (3 - 1) = -2/2 = -1 <== ur slope is -1
3 0
3 years ago
Describe the end behavior of a 14th degree polynomial with a positive leading coefficient.
kipiarov [429]
The function will enter the graph graph in the upper left hand region and exit in the upper right hand region and overall the graph will be concave upwards.  
For determining the end behavior of a polynomial, there is just 2 things to take notice of. 
1. Is the leading coefficient positive or negative? 
2. Is the degree of the polynomial odd or even?  
For odd ordered polynomials, the curve starts in either quadrant II or III, and ends in quadrant IV, or I. Basically, if it's positive, the curve enters the graph somewhere in the lower left hand region, and exits the graph in the upper right hand region. If the coefficient is negative, it enters in the upper left hand region, and exits in the lower right hand region.  
For even ordered polynomials, the graph is either concave upwards (positive leading coefficient) or concave downwards (negative leading coefficient).  
In this problem, 14 is an even number and since the coefficient is positive, the function will enter the graph graph in the upper left hand region and exit in the upper right hand region and overall the graph will be concave upwards.
4 0
3 years ago
What is 2 3/8 as a percent
Maru [420]
Percent means parts out of 100 so x%=x/100
convert to improper fraction for in x/y form so

2 and 3/8
2=16/8
2 and 3/8=16/8+3/8=19/8
convert bottom number to 100
one way is multiply whole thing by 100/100
19/8 times 100/100=1900/800
divide by 8
237.5/100=237.5%
that is precent
another way is divide completely
19/8=2.375
that is over 1 so convert to over 100 so
2.375/1 times 100/100=237.5/100=237.5%

answe ris 237.5%
6 0
3 years ago
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