How to write two hundred thirty-eight thousand, five hundred thirty-three using suymbols please help

(a) How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers?

Olegator [25]

Answer:

a) 67600000 possible plates

b) 19656000 possible plates where neither digit or letter are repeated

Step-by-step explanation:

The easiest way to solve this problem is seeing how many possibilities and there for each place in the license plate:

We have the following 7 places:

P1 - P2 - P3 - P4 - P5 - P6 - P7

The alphabet has 26 letters and there are 10 digits. So,

a) For each position there are the following possibilites

26 - 26 - 10 - 10 - 10 - 10 - 10

So in all, there are, 26*26*10*10*10*10*10 = 67600000 possible plates

b) Let's do the possibilites for each position again. Now, no letter or number can be repeated, so.

P1 is still 26. Now for P2, the letter in position P1 cannot be repeated, so there are only 25 possibilies. As for the digits, for P3, the first digit, there are still 10 possibilities. For P4, there are 9, since the digit in P3 cannot be repeated. For P5 there are 8, since P3 and P4 cannot be repeated... So there are the following number of possibilities:

26-25-10-9-8-7-6

In all, there are 26*25*10*9*8*7*6 = 19656000 possible plates where neither digit or letter are repeated.

5 0

3 years ago

28 children will attend the birthday party if each child gets 2 party favors and there are 14 favors in each box how many boxes

dezoksy [38]

Hello!

So if each box has 14 favors, what we need to do is add until we have enough, or more than enough for each of the children that will be at the party to get two of them.

So to start, 14+14=28. There are 28 children going, but each child must have two favors, not one. So, we simply add on two more boxes. 28+14=42, 42+14=56. So, we need 56 favors for each child to get two.

You could multiply to make this problem faster, but adding works out easier in my head. Sorry to drag it out!

So to make it simple, you need four boxes of favors for each child to get two.

Hope this helps, have an awesome day!

3 0

3 years ago

When is the equation <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx-7%7D%7Bx%5E2-4x-21%7D%20%3D%5Cfrac%7B1%7D%7Bx%2B3%7D" id="Tex

Vladimir79 [104]

Notice that

$x^2 - 4x - 21 = (x - 7) (x + 3)$

so on the left side, a factor of x - 7 in the numerator and denominator can cancel

$\dfrac{x-7}{x^2-4x-21} = \dfrac{x-7}{(x-7)(x+3)} = \dfrac1{x+3}$

But we can only cancel them out as long as x ≠ 7 (because otherwise we'd have the indeterminate form 0/0 on the left side).

So, the equation is true for all x except x = 7. In set notation,

$\{x \in \mathbb R \mid x \neq 7\}$

In interval notation,

$(-\infty,7)\cup(7,\infty)$

8 0

3 years ago

The cost of a movie ticket is increased by 15%. The old price was five dollars how much are they now?

Grace [21]

Answer:

5.75 is the answer.

$5.00 x 0.15=0.75

$5.00+0.75=$5.75

4 0

3 years ago

Please help thank you.

MA_775_DIABLO [31]

Question 1:

To start off this question, we can tell that this is a square because it has 4 right angles and 4 congruent sides.

A square has four parallel sides and 4 congruent sides, so a square is a rhombus and parallelogram.

A square has 4 right angles, so it's also a rectangle.

A square has 4 sides, so it's also a quadrilateral.

The first choice is your answer.

Question 2:

Not all quadrilaterals are rectangles, so A is incorrect.

Not all quadrilaterals are squares, so B is incorrect.

All rectangles are types of quadrilaterals, so C is correct.

Not all quadrilaterals are parallelograms, so D is incorrect.

Thus, C is your answer.

Question 3:

The first choice will not work because a rhombus will satisfy those conditions, and a rhombus is not always a square.

The second choice will work because only a square will satisfy that condition because only squares have 4 congruent sides along with equal diagonals.

Thus, the second choice is your answer.

Have an awesome day! :)

7 0

4 years ago