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Reil [10]
3 years ago
12

Can I simplest 48/30

Mathematics
2 answers:
ElenaW [278]3 years ago
8 0
Yes, 48/30 in simplest form is 8/5.
brilliants [131]3 years ago
7 0
Yes. it would be 1 and 18/30 or 1 9/15. it can also be 1 and 3/5 or 8/5
You might be interested in
How to solve the equation for the indicated variable: a-6[b-3(c-x)], for x.
Brilliant_brown [7]
6b-18c+18x is simplified for you


4 0
3 years ago
A 90 percent confidence interval for the slope of a regression line is determined to be (-0.181, 1.529). Which of the following
GenaCL600 [577]

Answer: a. The correlation coefficient of the data is positive.

Step-by-step explanation:

Estimated slope of sample regression line  = \dfrac{Upper \ limit + lower\ limit}{2}

Here , confidence interval :  (-0.181, 1.529)

Estimated slope of sample regression line  = \dfrac{-0.181+1.529}{2}

=\dfrac{1.348}{2}\\\\=0.674\ \ \ \ [\text{ positive}]

⇒Correlation coefficient(r) must be positive, So a. is true.

But, d. and e. are wrong(0.674 ≠ 0 or 1.348).

We cannot check residuals or its sum from confidence interval of slope of a regression line, so b is wrong.

We cannot say that scatterplot is linear as we cannot determine it from interval, so c. is wrong

So, the correct option : a. The correlation coefficient of the data is positive.

7 0
3 years ago
Why is it necessary to check that n modifyingabove p with caret greater than or equals 5np≥5 and n modifyingabove q with caret g
Nookie1986 [14]

Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if np \ge 5 and nq \ge 5 then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.

side note: q = 1-p is the complement of probability p

4 0
3 years ago
Gardening Mitch just finished planting radishes, cabbage, spinach, poas, and celery in his new garden. The garden is a circle wh
REY [17]
Answer: A= 424.5 meters^2

3 0
2 years ago
The data below show the weights, in pounds, of a group of professional football players.
monitta

Answer:

The Mean Absolute Deviation is 11

Step-by-step explanation:

Given:

325, 310, 289, 288, 285, 285, 285, 280, 280 and 273

Mean = 290

Required:

Calculate the Mean Absolute Deviation

Provided that we have the value of the mean to be 290 from the question; the following steps will calculate the Mean Absolute Deviation

Step 1: Subtract the mean weight from each weight

325 - 290 = 35

310 - 290 = 20

289 - 290 = -1

288 - 290 = -2

285 - 290 = -5

285 - 290 = -5

285 - 290 = -5

280 - 290 = -10

280 - 290 = -10

273 - 290 = -17

Step 2: Calculate the absolute value of the each results above

|35| = 35

|20| = 20

|-1| = 1

|-2| = 2

|-5| = 5

|-5| = 5

|-5| = 5

|-10| = 10

|-10| = 10

|-17| = 17

Step 3: Calculate the mean of the data above

Mean Absolute Value = \frac{35 + 20 + 1 + 2 + 5 + 5 + 5 + 10 + 10 + 17}{10}

Mean Absolute Value = \frac{110}{10}

Mean Absolute Value = 11

Hence, the Mean Absolute Deviation is 11

3 0
3 years ago
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