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Vikentia [17]
3 years ago
14

How many two digit numbers have one odd digit and one even digit?

Mathematics
2 answers:
sineoko [7]3 years ago
3 0

Disclaimer: I'm assuming that you want any of the digit to be even, and the other to be odd, and not that you mean "the first digit odd and the second even".

You can note that there are 100 numbers with two digits, from 0 (which you can read as 00) to 99.

Now, we're only interested in those who have one odd digit and one even digit. Let's see if we can find a pattern in the numbers we do or don't want to include in our list:

Consider the numbers from 00 to 09. Since the first digit is zero, which is even, the unit digit must be odd. So, we must exlude the following numbers:

00, 02, 04, 06, 08

while the following numbers are ok:

01, 03, 05, 07, 09

Now, consider the numbers from 10 to 19. This time the first digit is one, which is odd, so the unit digit must be even. Now, we must exlude the following numbers:

11, 13, 15, 17, 19

while the following numbers are ok:

10, 12, 14, 16, 18

Can you see the pattern? If you consider any 10 consecutive numbers with the same first digit, you will always keep half of the numbers and discard the other half. In fact, if the first digit e is even, you will keep e1, e3, e5, e7 \text{ and } e9, while you will discard e0, e2, e4, e6 \text{ and } e8.

If, instead, the first digit o is odd, your choice will be the opposite: you will discard o1, o3, o5, o7 \text{ and } o9, while you will keep o0, o2, o4, o6 \text{ and } o8.

So, we've just proven that exactly half of the numbers with two digits have one odd digit and one even digit. We observed at the beginning that there are 100 two-digits number in total, so there are 50 numbers with one even digit and one odd digit.

sergiy2304 [10]3 years ago
3 0

Answer:

45 is the answer

Step-by-step explanation:

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Plant Type Number tomato 8 squash 4 cucumber 16 rosemary 22 he The table above shows the number of each type of plant in Robert'
kramer

Answer:

C. 25:4

Step-by-step explanation:

Total number of plants in Robert's garden = 8 + 4 + 16 + 22

= 50

Number of tomato plants in Robert's garden = 8

Ratio of total number of plants to the number of tomato plants = 50 : 8

Simplify

50 : 8 = 25:4

4 0
3 years ago
The area of a playground is 20 square yards. The length of the playground is 5 times longer than its width. Find the length and
jok3333 [9.3K]
The answer is <span>length = 10 yd, width = 2 yd
proof 
</span><span>The length of the playground is 5 times longer than its width, it means
l=5w, but 10= 2x5 so we can write l= 5w, 
and Area = 20 =l x w = (5x2) x 2 =20</span>
6 0
3 years ago
What is the term-to-term rule for this sequence?<br> 64, 32, 16, 8, 4
just olya [345]

Answer:

a n=1/2 an - 1

Step-by-step explanation:

64,32,16,8,4. is a geometric sequence. this is a sequence where, to get from one term to the next, there is a number that you have to multiply it by. this is the same for all parts of the sequence.

4 0
3 years ago
each of the letters of the word ALABAMA is written on a piece of paper and then put into a bag. A piece of paper is drawn at ran
Yanka [14]

Answer:

P(A) = 0.57 (to the nearest hundredth)

Step-by-step explanation:

There are 7 letters in total

4 of them are the letter A

⇒ P(A) = 4/7

           = 0.5714285714...

           = 0.57 (to the nearest hundredth)

5 0
3 years ago
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f)  ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

7 0
3 years ago
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