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arsen [322]
3 years ago
7

A babysitting job pays $4 per hour.

Mathematics
1 answer:
Anika [276]3 years ago
4 0

Answer:

1

2

$8

3

$12

4

$16

Step-by-step explanation:

You might be interested in
7/16 in decimal form​
neonofarm [45]

Answer:

0.4375

Step-by-step explanation:

8 0
3 years ago
Please answer ill pay 50
Sergio039 [100]

Answer:

1:12

2:-32

3: -35

4: -10

5: 24

6: 90

7: 41.  I used the properties of multiplication to find the product by multiplying 41•1=41 and the 2 negatives cancel each other out.

8: the product is -80. What she could've done is forgot to add the negative sign.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Play help due tomorrow
IgorC [24]
D. 42/8

You can get this by adding the blueberry and banana amounts together. Then multiply by 6. 
5 0
3 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
4 years ago
Which number is the answer
const2013 [10]

Answer:

1

Step-by-step explanation:

32-24 gives you 8

So she only needs one.

7 0
3 years ago
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