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sashaice [31]
2 years ago
11

Please help will wark brainliest

Mathematics
1 answer:
Dima020 [189]2 years ago
3 0
0.7 divided by 0.35 = 2 is the correct answer :))
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Which of the binomials below is a factor of this trinomial x^2+4x+4
dmitriy555 [2]
X² + 4x + 4
x            2
x             2

(x + 2) (x + 2) 


x + 2 should be your answer

binomial is having 2 terms

hope this helps
4 0
3 years ago
Read 2 more answers
The ratio of Boys to Girls in Mr. Asante Class is 2 to 3. There are 18 girls in the class. What is the total number of students
ladessa [460]

Answer:

30

Step-by-step explanation:

2/3 = x/18

2*18 = 3x

36 = 3x

x=12

so there are 12 boys

12+18=30

7 0
3 years ago
Which of the following expressions can be used to find the perimeter of a rectangle?
morpeh [17]

Answer:

2l + 2w. l+l+w+w. l+l+2w. 2(l+w)

5 0
3 years ago
A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements are centimeters on a pain sca
blondinia [14]

The confidence interval for mean of the "before-after" differences is \fbox{(-0.4037,1.8037)}

Further explanation:

Find the difference between the before pain and the after pain.

Difference = before-after

Kindly refer to the Table for the difference of between the before and after pain.

Sum of difference = 5.6

Total number of observation = 8

Mean of difference = 0.7

Sample standard deviation s = 1.3201

Level of significance = 5\%

Formula for confidence interval = \left( \bar{X} \pm t_{n-1, \frac{\alpha}{2}\%} \frac{s}{\sqrt{n}} \right)

confidence interval = \left( 0.7 \pm t_{8-1, \frac{5}{2}\%} \frac{1.3201}{\sqrt{8}} \right)

confidence interval = \left( 0.7 \pm t_{7, \frac{5}{2}\%} \frac{1.3201}{\sqrt{8}} \right)

From the t-table.

The value of t_{7, \frac{5}{2}\%=2.365

Confidence interval = ( 0.7 \pm 2.365}\times \frac{1.3201}{\sqrt{8}}) \right)

Confidence interval = \left( 0.7 - 2.365}\times \0.4667,0.7 + 2.365}\times \0.4667) \right

Confidence interval = (0.7-1.1037,0.7+1.1037)

Confidence interval = \fbox{(-0.4037,1.8037)}

The 95\% confidence interval tells us about that 95\% chances of the true mean or population mean lies in the interval.

Yes, the hypnotism appear to be effective in reducing pain as confidence interval include includes the positive deviation from the mean.

Learn More:

1. Learn more about equation of circle brainly.com/question/1506955

2. Learn more about line segments brainly.com/question/909890  

Answer Details:

Grade: College Statistics

Subject: Mathematics

Chapter: Confidence Interval

Keywords:

Probability, Statistics, Speed dating, Females rating, Confidence interval, t-test, Level of significance , Normal distribution, Central Limit Theorem, t-table, Population mean, Sample mean, Standard deviation, Symmetric, Variance.

7 0
3 years ago
Read 2 more answers
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
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