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ASHA 777 [7]
2 years ago
5

Save and

Mathematics
1 answer:
Darya [45]2 years ago
7 0

Answer: 6/5

Step-by-step explanation:

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Plz help meeeeeeee <br>I need to turn this in today<br>plzzz!!!!!!​
Katarina [22]

Answer:



Step-by-step explanation:

Step 1 Answer: Yes, 3

Step 2 Answer: Yes, 5

Step 3 Answer: No

8 0
2 years ago
Read 2 more answers
X + y = 8 and x + 2y =14
cricket20 [7]

Answer:

x = 2

y=6

Step-by-step explanation:

6 + 2 = 8

2 + 6 x 2 = 14

Hope this helps

Have a blessed day!

3 0
3 years ago
A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a coll
xxMikexx [17]

Answer: D

H0: μ=522

H1: μ>522

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

So, for this case;

The null hypothesis is that the mean score equals to 522

H0: μ=522

The alternative hypothesis is that the mean score is greater than 522.

H1: μ>522

8 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
3 years ago
Find the value of X in the triangle shown.
zalisa [80]
X = 12

Use the Pythagorean Theorem.
13^2 = 5^2 + x^2
169 = 25 + x^2
Subtract 25 from both sides.
144 = x^2
Take the square root.
12 = x
7 0
2 years ago
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