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Bond [772]
3 years ago
12

11/12 - ___ + 1/4 = 2/3

Mathematics
1 answer:
kondaur [170]3 years ago
4 0
7/6-____=2/3
2/3-7/6 =-x
-1/2=-x
The negatives cancel out so the blank = 1/2
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Five observations taken for two variables follow.
skad [1K]

Answer:

-60 ; -0.9689

Step-by-step explanation:

Given the data:

xi 4 6 11 3 16

yi 50 50 40 60 30

The scatter diagram indicates a linear negative relationship between x and y. Thia is indicated by the negative direction of the trend line.

Sample covariance formula:

Σ(x-xi)(y-yi) / n - 1

Using calculator : sample covariance = - 60

The variables have negative covariance meaning an increase in one variable leads to a corresponding decrease in the other.

Sample correlation Coefficient : -0.9689

This shows that a very strong negative correlation exists between both variables. Because the correlation coefficient value is very close to - 1.

5 0
3 years ago
Can anybody help me with this math problems?
Ivan

Answer:

All you have to do is replace the X with the number on the table

Step-by-step explanation:

so the first number is -3 so 3×-3+2

that equals -7 so put -7 as the first number under y. can continue like that

3 0
3 years ago
Read 2 more answers
A 90% confidence interval for the mean percentage of airline reservations being canceled on the day of the flight is (1.1%, 3.4%
Anton [14]
The point estimate would be 2.25%.

Confidence intervals are centered around a point estimate; that is, the point estimate is in the very middle of the confidence interval.  We can find the point estimate by averaging both ends of the confidence interval together:

(1.1+3.4)/2 = 2.25.
6 0
4 years ago
A club swimming pool is 21 ft wide and 33 ft long. the club members want an exposed aggregate border in a strip of uniform width
vovangra [49]
First compute the perimeter of the pool:
(21+33)*2=108 foot. 
Since the material cover 232 ft^2, then it is sufficient to divide the above number over the length of the border like this:
232/108=2.1

The trip can be 2.1 ft in wide. 
5 0
3 years ago
2. Lab groups of three are to be randomly formed (without replacement) from a class that contains five engineers and four non-en
Anna11 [10]

Answer:

The number of different lab groups possible is 84.

Step-by-step explanation:

<u>Given</u>:

A class consists of 5 engineers and 4 non-engineers.

A lab groups of 3 are to be formed of these 9 students.

The problem can be solved using combinations.

Combinations is the number of ways to select <em>k</em> items from a group of <em>n</em> items without replacement. The order of the arrangement does not matter in combinations.

The combination of <em>k</em> items from <em>n</em> items is: {n\choose k}=\frac{n!}{k!(n-k)!}

Compute the number of different lab groups possible as follows:

The number of ways of selecting 3 students from 9 is = {n\choose k}={9\choose 3}

                                                                                         =\frac{9!}{3!(9 - 3)!}\\=\frac{9!}{3!\times 6!}\\=\frac{362880}{6\times720}\\ =84

Thus, the number of different lab groups possible is 84.

8 0
3 years ago
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