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237 what is the place value of 3 Place value for integers with no decimals from right to left is: 7 is the ones digit 3 is the [B]tens digit[/B]

756.218 which digit is in the thousands place Starting from the right of the decimal, we have the tenths place, hundreds place, thousands place. So [B]8[/B] is the thousands place

A 3-digit security code can use the numbers 0-9. How many possible combinations are there if the numbers can be repeated [0-9] * [0-9] * [0-9] 10 * 10 * 10 = [B]1,000 combinations[/B]

A bag of marbles is said to contain 50 marbles to the nearest ten. What is the greatest number of marbles that could be in the bag and what is the least number of marbles that could be in the bag The key word in this problem is [I][B]nearest ten[/B][/I]. The nearest ten below 50 starts at 45. Why? Because the last digit is 5. At 5, we round up to the nearest ten. Therefore, the least number of marbles in the bag is 45 since it rounds up to 50 for the nearest ten The greatest number above 50 rounded to the nearest ten is 54, because less than 5 on the last digit means we round down. Therefore, the greatest number of marbles in the bag is 54 since it rounds down to 50 when the last digit is less than 5 Answer: {[B]45, 54} [MEDIA=youtube]-cl_OHA8-yc[/MEDIA][/B]

a licence plate that has 3 numbers from 0 to 9 followed by 2 letters How many license plate combinations can we form? We multiply as follows: [LIST] [*][0-9] = 10 possible digits (D) [*]A-Z = 26 possible letters (L) [/LIST] The problem asks for this: DDDLL So we have: 10 * 10 * 10 * 26 * 26 = [B]676,000[/B] plates

A license plate is made up of 2 letter and 3 single digit numbers. There are 26 letters (A-Z). And there are 10 single digit numbers [0-9]. So our total combinations are: Letter - Letter - Number - Number - Number 26 * 26 * 10 * 10 * 10 = [B]676,000[/B]

A numerical pass code is required to open a car door. The pass code is five digits long and uses the digits 0-9. Numbers may be repeated in the pass code. How many different pass codes exist? 0-9 is 10 digits. Since digits can repeat, we use the fundamental rule of counting to get: 10 * 10 * 10 * 10 * 10 = [B]100,000 different pass codes[/B]

A three digit number, if the digits are unique [LIST=1] [*]For our first digit, we can start with anything but 0. So we have 9 options [*]For our second digit, we can use anything but 9 since we want to be unique. So we have 9 options [*]For our last digit, we can use anything but the first and second digit. So we have 10 - 2 = 8 options [/LIST] Our total 3 digit numbers with all digits unique is found by the fundamental rule of counting: 9 * 9 * 8 = [B]648 possible 3 digit numbers[/B]

Chicken is on sale for $3.90 per pound. If Ms.Gelllar buys 2.25 pounds of chicken, how much will she spend? round to the nearest penny and show your work Total spend = Cost per pound * Number of pounds Total spend = $3.90 * 2.25 pounds Total spend = [B]$8.78[/B] (rounded to 2 digits)

Free Compound Interest and Annuity Table Calculator - Given an interest rate (i), number of periods to display (n), and number of digits to round (r), this calculator produces a compound interest table. It shows the values for the following 4 compound interest annuity functions from time 1 to (n) rounded to (r) digits:
vⁿ
d
(1 + i)ⁿ
a_n|
s_n|
ä_n|i
s_n|i
Force of Interest δⁿ

Derek must choose a 4 digit PIN. Each Digit can be chosen from 0 to 9. Derek does not want to reuse any digits. He also only wants an even number that begins with 5. How many possible PINS could he choose from? [LIST=1] [*]First digit must begin with 5. So we have 1 choice [*]We subtract 1 possible digit from digit 3 to have 8 - 1 = 7 possible digits [*]This digit can be anything other than 5 and the even number in the next step. So we have 0-9 is 10 digits - 2 = 8 possible digits [*]Last digit must end in 0, 2, 4, 6, 8 to be even. So we have 5 choices [/LIST] Our total choices from digits 1-4 are found by multiplying each possible digit choice: 1 * 7 * 8 * 5 = [B]280 possible PINS[/B]

Free Dewey Decimal System Classification Calculator - Given a 3 digit code, this will determine the class, division, and section of the library book using the Dewey Decimal System.

Free Digit Problems Calculator - Determines how many (n) digit numbers can be formed based on a variety of criteria.

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2 rules. If either of them passes, then the number is divisible by 11: [LIST=1] [*]Sum of the odd digits - Sum of the even digits is divisible by 11 [*]Sum of the odd digits - Sum of the even digits = 0 (Ex. 121) [/LIST] [MEDIA=youtube]WpV87es0WAU[/MEDIA]

Check out this pattern: 2^1= 2 2^2= 4 2^3 = 8 2^4= 16 2^5 = 32 2^6 = 64 2^7 = 128 2^8 = 256 The last digit repeats itself in blocks of 4 2, 4, 8, 6 We want to know what is the largest number in 1, 2, 3, 4 that divides 2020 without a remainder. LEt's start with 4 and work backwards. 2020/4 = 505 Ever power of 2^4(n) ends in 6, so our answer is [B]6 [MEDIA=youtube]6uX5gwb1jdY[/MEDIA][/B]

Find the last digit of 4^2081 no calculator 4^1= 4 4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024 4^6 = 4096 Notice this pattern alternates between odd exponent powers with the result ending in 4 and even exponent powers with the result ending in 6. Since 2081 is odd, the answer is [B]4. [MEDIA=youtube]ueBWAW4XW4Q[/MEDIA][/B]

Consider the first 8 calculations of 7 to an exponent: [LIST] [*]7^1 = 7 [*]7^2 = 49 [*]7^3 = 343 [*]7^4 = 2,401 [*]7^5 = 16,807 [*]7^6 = 117,649 [*]7^7 = 823,543 [*]7^8 = 5,764,801 [/LIST] Take a look at the last digit of the first 8 calculations: 7, 9, 3, 1, 7, 9, 3, 1 The 7, 9, 3, 1 repeats through infinity. So every factor of 4, the cycle of 7, 9, 3, 1 restarts. Counting backwards from 2013, we know that 2012 is the largest number divisible by 4: 7^2013 = 7^2012 * 7^1 The cycle starts over after 2012. Which means the last digit of 7^2013 = [B]7 [MEDIA=youtube]Z157jj8R7Yc[/MEDIA][/B]

Find the odd number less than 100 that is divisible by 9, and when divided by 10 has a remainder of 7. From our [URL='http://www.mathcelebrity.com/divisibility.php?num=120&pl=Divisibility']divisibility calculator[/URL], we see a number is divisible by 9 if the sum of its digits is divisible by 9. Starting from 1 to 99, we find all numbers with a digit sum of 9. This would be digits with 0 and 9, 1 and 8, 2 and 7, 3 and 6, and 4 and 5. 9 18 27 36 45 54 63 72 81 90 Now remove even numbers since the problem asks for odd numbers 9 27 45 63 81 Now, divide each number by 10, and find the remainder 9/10 = 0 [URL='http://www.mathcelebrity.com/modulus.php?num=27mod10&pl=Calculate+Modulus']27/10[/URL] = 2 R 7 We stop here. [B]27[/B] is an odd number, less than 100, with a remainder of 7 when divided by 10.

Ok. To go further in this equation. It reads: ...How many did he originally take to the event? Multiply the answer by 3 and reverse the digits. This will give you the answer for ACH in the coordinates. Does that make sense to reverse 303? :-/ Thank you for your help!!

Let me post the whole equation paragraph: Brainteaser # 1: Answer for ACH A fellow geocacher decided that he would try to sell some hand-made walking sticks at the local geocaching picnic event. In the first hour, he sold one-half of his sticks, plus one-half of a stick. The next hour, he sold one-third of his remaining sticks plus one-third of a stick. In the third hour, he sold one-fourth of what he had left, plus three-fourths of a stick. The last hour, he sold one-fifth of the remaining sticks, plus one-fifth of a stick. He did not cut up any sticks to make these sales. He returned home with 19 sticks. How many did he originally take to the event? Multiply the answer by 3 and reverse the digits. This will give you the answer for ACH in the coordinates. Make sure to multiply and reverse the digits. What would the answer be?

Multiply the answer by 3: 101 * 3 = 303 Reverse the digits: 303 reversed is a palindrome, so it's still [B]303[/B].

Exam answers and Study Guide for the Google Digital Sales Exam

How many license plates can be made consisting of 3 letters followed by 2 digits There are 26 letters A-Z and 10 digits 0-9. We have: 26 * 26 * 26 * 10 * 10 = [B]1,757,600 license plates[/B]

In 203.46, which digit is in the hundredths place? Moving from the right of the decimal place, we have : tenths place = 3 hundredths place = [B]6[/B]

In 32, what is the value of 2 For place value, starting from the right decimal with no decimals, we have: tens, ones 3 is the tens digit 2 is the ones digit 32 = 3 * 10 + 2 * 1 Which means 2 is the [B]ones digit[/B]

Is someone has $1,000,000 in base 2, how much money does she have in base 10? 1 is in 7th digit place, so we raise it to the 6th power: [URL='https://www.mathcelebrity.com/powersq.php?sqconst=+6&num=2%5E6&pl=Calculate']1 * 2^6 [/URL]= [B]64[/B]

James Bond has a secret code. The code is 3 digits long and less than 160. The digits add to 10. What is his secret code? less than 160 means 0 to 159 Working backwards with 1 in the hundreds place and 5 in the 10's place, we see that 1 + 5 + 4 = 10 [B]154[/B]

Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What could N be? Is there more than one answer? For example, for 23 P(23) = 6 and S(23) = 5, but 23 could not be the N that we want since 23 <> 5 + 6 Let t = tens digit and o = ones digit P(n) = to S(n) = t + o P(n) + S(n) = to + t + o N = 10t + o Set them equal to each other N = P(N) + S(N) 10t + o = to + t + o o's cancel, so we have 10t = to + t Subtract t from each side, we have 9t = to Divide each side by t o = 9 So any two-digit number with 9 as the ones digit will work: [B]{19,29,39,49,59,69,79,89,99}[/B]

license plate with 4 letter combinations and 3 number combinations There are 26 total letters and 10 digits [0-9]. We have 26 C 4 * 10 C 3. [URL='http://www.mathcelebrity.com/permutation.php?num=26&den=4&pl=Combinations']26 C 4[/URL] = 14,950 [URL='http://www.mathcelebrity.com/permutation.php?num=10&den=3&pl=Combinations']10 C 3[/URL] = 120 Total license plate combinations: 14,950 * 120 = [B]1,794,000[/B]

License plates are made using 3 letters followed by 2 digits. How many plates can be made if repetition of letters and digits is allowed. We have 26 possible letters and 10 possible digits 0-9. Since repetition is allowed, we have: 26 * 26 * 26 * 10 * 10 = [B]1,757,600 possible license plates[/B]

License plates are made using 3 letters followed by 2 digits. How many plates can be made if repetition of letters and digits is allowed We have 26 letters A-Z and 10 possible digits 0-9. Using the fundamental rule of counting, we have: 26 * 26 * 26 * 10 * 10 = [B]1,757,600 possible choices[/B]

License plates are made using 3 letters followed by 3 digits. How many plates can be made of repetition of letters and digits is allowed We have 26 letters in the alphabet We have 10 digits [0-9] The problem asks for the following license plate scenario of Letters (L) and Digits (D) LLLDDD The number of plates we can make using L = 26 and D = 10 using the fundamental rule of counting is: Number of License Plates = 26 * 26 * 26 * 10 * 10 * 10 Number of License Plates = [B]17,576,000[/B]

[U]53 x 11[/U] Write 3 as the right most digit in 53 3 5 + 3 = 8. Write that to the left of 3 83 Then write the 5 as the left most digit 583 [U]35 x 11[/U] Write 5 as the right most digit in 35 5 3 + 5 = 8. Write that to the left of 5 85 Then write the 3 as the left most digit 385 [U]57 x 11[/U] Write the 7 in 57 as the leftmost digit 7 5 + 7 = 12 Write the 2 in 12 as the next digit 27 Since our last sum was 10 or greater, we add 1 to the 5 in 57 to get 6 627 [U]91 x 11[/U] Write the 1 in 91 as the leftmost digit 1 9 + 1 = 10 Write the 0 in 10 as the next digit 01 Since our last sum was 10 or greater, we add 1 to the 9 in 91 to get 10 1001 [MEDIA=youtube]Uzfj57veazA[/MEDIA]

Free Natural Logarithm Table Calculator - Generates a natural logarithm table for the first (n) numbers rounded to (r) digits

Get a free pi coin. Use the link below: I am sending you 1?! Pi is a new digital currency developed by Stanford PhDs, with over 47 million members worldwide. To claim your Pi, follow this link [URL]https://minepi.com/mathcelebrity[/URL] and use my username (mathcelebrity) as your invitation code.--

Serial numbers for a product are to be using 3 letters followed by 4 digits. The letters are to be taken from the first 8 letters of the alphabet with no repeats. The digits are taken from numbers 0-9 with no repeats. How many serial numbers can be generated The serial number is organized with letters (L) and digits (D) like this: LLLDDDD Here's how we get the serial number: [LIST=1] [*]The first letter can be any of 8 letters A-H [*]The second letter can be any 7 of 8 letters A-H [*]The third letter can be any 6 of 8 letters A-H [*]The fourth digit can be any of 10 digits 0-9 [*]The fifth digit can be any 9 of 10 digits 0-9 [*]The sixth digit can be any 8 of 10 digits 0-9 [*]The seventh digit can be any 7 of 10 digits 0-9 [/LIST] We multiply all possibilities: 8 * 7 * 6 * 10 * 9 * 8 * 7 [B]1,693,440[/B]

Serial numbers for a product are to made using 4 letters followed by 4 numbers. If the letters are to be taken from the first 5 letters of the alphabet with repeats possible and the numbers are taken from the digits 0 through 9 with no repeats, how many serial numbers can be generated? First 5 letters of the alphabet are {A, B, C, D, E} The 4 letters can be chosen as possible: 5 * 5 * 5 * 5 The number are not repeatable, so the 4 numbers can be chosen as: 10 * 9 * 8 * 7 since we have one less choice with each pick Grouping letters and numbers together, we have the following serial number combinations: 5 * 5 * 5 * 5 * 10 * 9 * 8 * 7 = [B]3,150,000[/B]

Set C is the set of two-digit even numbers greater than 72 that do not contain the digit 8. First, two-digit numbers mean anything less than 100. Let's, list out our two-digit even numbers greater than 72 but less than 100. C = {74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98} The problem asks for numbers that do not contain the digit 8. Let's remove those numbers from the list. C = {74, 76, [S]78[/S], [S]80, 82, 84, 86, 88[/S], 90, 92, 94, 96, [S]98[/S]} [B]C = {74, 76, 90, 92, 94, 96} [MEDIA=youtube]_O6nXX0V4zo[/MEDIA][/B]

[U]Two digit Numbers less than 56:[/U] {10, 11, 12, ..., 55} [U]Two Digit Even Numbers of that Set:[/U] {10, 12, 14, ..., 54} [U]Two Digit Even numbers Divisible by 5[/U] [B]C = {10, 20, 30, 40, 50}[/B] [I]Note: Even means you can divide it by 2 with no remainder. Divisible by 5 means the number ends in 5 or 0. Since it is even numbers only, end in 0. [MEDIA=youtube]aQKLVxIB-p4[/MEDIA][/I]

Set D is the set of two-digit even numbers less than 67 that are divisible by 5 two-digit numbers start at 10. Divisible by 5 means the last digit is either 0 or 5. But even numbers don't end in 5, so we take the two-digit numbers ending in 0: D = {[B]10, 20, 30, 40, 50, 60}[/B]

Set of 2 digit even numbers less than 40 Knowns and givens: [LIST] [*]2 digit numbers start at 10 [*]Less than 40 means we do not include 40 [*]Even numbers are divisible by 2 [/LIST] [B]{10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38}[/B]

Free Square Root Table Calculator - Generates a square root table for the first (n) numbers rounded to (r) digits

Subtracting 9s shortcut Add the digits of the larger number [LIST] [*]10 - 9 = 1 + 0 = 1 [*]11 - 9 = 1 + 1 = 2 [*]12 - 9 = 1 + 2 = 3 [*]13 - 9 = 1 + 3 = 4 [*]14 - 9 = 1 + 4= 5 [*]15 - 9=. 1 + 5 = 6 [*]16 - 9 = 1 + 6= 7 [*]17 - 9= 1 + 7 = 8 [*]18 - 9 = 1 + 8 = 9 [*]19 - 9 = 1 + 9 = 10 [/LIST] [MEDIA=youtube]YOHcJ6UG1D8[/MEDIA]

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Suppose you secured your phone using a passcode. Later, you realized that you forgot the 6-digit code. You only remembered that the code contains the digits 1, 2,3, 4,5 and 6. How many possible codes can there be? 6 possible digits, 1-6 and the code is 6-digits long. So we have: 6 * 6 * 6 * 6 * 6 * 6 = 6^6 = [B]46,656 possible codes[/B]

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford's Law. For example, the following distribution represents the first digits in 231 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer. Digit, Probability 1, 0.301 2, 0.176 3, 0.125 4, 0.097 5, 0.079 6, 0.067 7, 0.058 8, 0.051 9, 0.046 [B][U]Fradulent Checks[/U][/B] Digit, Frequency 1, 36 2, 32 3, 45 4, 20 5, 24 6, 36 7, 15 8, 16 9, 7 Complete parts (a) and (b). (a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?
Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.

The ones digit of a two-digit number is three, while the tens digit is four. We write this as tens digit ones digit: [B]43[/B]

The place value of 3 in 16.534 is We [URL='https://www.mathcelebrity.com/placevalue.php?num=16.534&pl=Show+Place+Value']type in 16.534 into our search engine, choose place value[/URL], and we get: 3 is the [B]hundredths digit[/B]

The sum of the digits of a 2 digit number is 10. The value of the number is four more than 15 times the unit digit. Find the number. Let the digits be (x)(y) where t is the tens digit, and o is the ones digit. We're given: [LIST=1] [*]x + y = 10 [*]10x+ y = 15y + 4 [/LIST] Simplifying Equation (2) by subtracting y from each side, we get: 10x = 14y + 4 Rearranging equation (1), we get: x = 10 - y Substitute this into simplified equation (2): 10(10 - y) = 14y + 4 100 - 10y = 14y + 4 [URL='https://www.mathcelebrity.com/1unk.php?num=100-10y%3D14y%2B4&pl=Solve']Typing this equation into our search engine[/URL], we get: y = 4 Plug this into rearranged equation (1), we get: x = 10 - 4 x = 6 So our number xy is [B]64[/B]. Let's check our work against equation (1): 6 + 4 ? 10 10 = 10 Let's check our work against equation (2): 10(6)+ 4 ? 15(4) + 4 60 + 4 ? 60 + 4 64 = 64

The sum of the digits of a certain two-digit number is 16. Reversing its digits increases the number by 18. What is the number? Let x and (16-x) represent the original ten and units digits respectively Reversing its digits increases the number by 18 Set up the relational equation [10x + (16-x)] + 18 = 10(16 - x) + x Solving for x 9x + 34 = 160 - 9x Combine like terms 18x = 126 Divide each side of the equation by 18 18x/18 = 126/18 x = 7 So 16 - x = 16 - 7 = 9 The first number is 79, the other number is 97. and 97 - 79 = 18 so we match up. The number in our answer is [B]79[/B]

To create an entry code, you must first choose 2 letters and then, 4 single-digit numbers. How many different entry codes can you create? List total combinations using the product of all possibilities: 26 letters (A - Z) * 26 letters (A - Z) * 10 digits (0-9) * 10 digits (0-9) * 10 digits (0-9) * 10 digits (0-9) [B]6,760,000 entry codes [MEDIA=youtube]Y23EGnVuU7I[/MEDIA][/B]

what does the digit 7 in 65.47 stand for To the right of the decimal place, moving left to right, we have: 4 is the tenths place [B]7 is the hundredths place[/B]

Write in set builder form {all possible numbers formed by any two of the digits 1 2 5} With 3 numbers, we got [URL='https://www.mathcelebrity.com/factorial.php?num=3!&pl=Calculate+factorial']3! = 6[/URL] possible numbers formed by the two digits [LIST=1] [*]12 [*]15 [*]21 [*]25 [*]51 [*]52 [/LIST] In set builder notation, we write this as: {x : x ? {12, 15, 21, 25, 51, 52}) x such that x is a element of the set {12, 15, 21, 25, 51, 52}