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Eddi Din [679]
4 years ago
14

Lana wants to buy a books

Mathematics
2 answers:
bija089 [108]4 years ago
4 0
6 one dollar bills
9 dimes
1nickel

daser333 [38]4 years ago
4 0
$1 9 dimes 1 nickel i think so
You might be interested in
Write a word problem that can be solved by multiplying a whole number and a fraction. Include the solution.
adelina 88 [10]

Answer:

Sheila had 24 cupcakes, she gave 1/3 of them to Caleb. How many cupcakes did she give to Caleb?

Step-by-step explanation:

24 x 1/3 = 8

She gave 8 cupcakes to Caleb.

8 0
2 years ago
Solve the inequalty 2 2/5 < b-8/15
mr_godi [17]

Answer:

b>44

Step-by-step explanation:

12/5<b-8/15

switch sides b-8/15>12/5

multiply both sides by 15  15(b-8)/15>12*15/5

simplify b-8>36

add 8 to both sides b-8+8>36+8

simplify

b>44

4 0
4 years ago
Show the following conjectures are false by finding a counterexample.
Julli [10]

''All prime numbers are odd."

That is incorrect because if we consider the smallest odd number, 2, is an even number. We can say that except 2, all prime numbers are odd.

14. section is wrong. For example, both numbers can be negative, so when we multiple these two, it will equals a positive value.

3 0
3 years ago
Determine how many different computer passwords are possible if (a) digits and letter so can be repeated and (b) digits and lett
Lerok [7]

Answer:

a) 1,188,137,600 different passwords.

b) 710,424,000 different passwords.

Step-by-step explanation:

We have a code of 2 digits followed by 5 letters.

First, the total number of digits is 10

The total number of letters is 26.

Then:

a) Digits and letters can be repeated.

Here we need to count the number of options for each selection.

For the first digit, we have 10 options.

For the second digit, we have 10 options.

For the first letter, we have 26 options.

For the second letter, we have 26 options.

For the third letter, we have 26 options.

For the fourth letter, we have 26 options.

For the fifth letter, we have 26 options.

The total number of combinations will be equal to the product of the number of options. We get:

Combinations = 10*10*26*26*26*26*26 = (10^2)*(26^5) = 1,188,137,600

This means that we have  1,188,137,600 different possible passwords.

b) Digits and letters can not be repeated.

We start in the same way as above:

For the first digit, we have 10 options.

For the second digit, we have 9 options, because one is already used.

For the first letter, we have 26 options.

For the second letter, we have 25 options, because one letter is already used.

For the third letter, we have 24 options, because 2 letters are already used.

For the fourth letter, we have 23 options, because 3 letters are already used.

For the fifth letter, we have 22 options, because 4 letters are already used.

Then the total number of combinations is:

Combinations = 10*9*26*25*24*23*22 = 710,424,000

So if we can not repeat digits nor letters, we can make 710,424,000 different passwords,

8 0
3 years ago
Would $105.00 be correct? With the equation of Y=x(15)?
Vilka [71]

Answer:

$15.00h=N

$15.00(7)=N

$105.00=N

N=$105.00

So yes you are correct

Hope that helped

4 0
3 years ago
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