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Anvisha [2.4K]
3 years ago
7

A photograph was enlarged to a width of 15 inches. If the scale factor was 3 /2 , what was the width of the original photograph?

Mathematics
1 answer:
andre [41]3 years ago
6 0
If the width is 15, and the scale factor from width to length is 3:2, then you will have to divide 15 by 3 to get your original width.
15/3 = 5.
5 is your original width.
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Is 1,341,491 divisible by 9?
zheka24 [161]

Answer:

no

Step-by-step explanation:

you would get this: 149054.555556

8 0
2 years ago
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Carly buys a gold ring priced at $273.00 if the sale tax is 10 % , how much will carly pay
Damm [24]

Hello!

<h2>Answer:</h2>

Carla will pay $300.30 for the gold ring.

<h2>Explanation:</h2>

Sales tax is added on to the total cost of the item. We need to figure out how much tax is on the ring.

To do that, we must calculate 10% (or 0.10) of $273.00.

273 × 0.10 = 27.3

Now, add the tax to the cost of the ring.

273 + 27.3 = 300.3

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3 years ago
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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
3 years ago
An investigative bureau uses a laboratory method to match the lead in a bullet found at a crime scene with unexpended lead cartr
Basile [38]

Answer / Step-by-step explanation:

In this exercise, we know that the agency collected 1,829 bullets which according to the agency, all came from different source.

The number of pair of bullet is :

         (1,829) = 1, 829 ! / 2! ( 1,829 - 2 ) = 1, 829 x 1,828 x 1827 ! / 2 x 1 x 1,827

          (   2   )

                   = 6, 108,413,724 / 3654

                   = 1671,706.

Also, we know that the agency found 658 matches. According to this, the chance of false positive ( that is, the agencies find a match but the bullets are from two different source ) is

   P (False positive ) = 658 / 671,706

                                   = 0.000979.

The probability is small to an extent, therefore, we should have confidence in the agency forensic report and evidence

6 0
3 years ago
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