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3 years ago

HELP IF RIGHT WILL MARK BRIANLIEST!! NO LINKS!!!!!!!

Mathematics

2 answers:

Brrunno [24]3 years ago

8 0

Answer:

14 is the simplified answer, but the whole answer is 17 + -18/6

Step-by-step explanation:

Divide

−

18 by 6. h (−18) = 17 − 3 Subtract 3 from 17. h (−18)= 14

The final answer is 14.

topjm [15]3 years ago

3 0

Yup there correct ^^^^ also Calculate84 on App Store life saver

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Identify the usual level of measurement for each of the following A. year in school B. IQ scores C. life expectancy D. fatigue E

Mashutka [201]

Answer:

The answer is C

Step-by-step explanation:

4 0

3 years ago

Evaluate the expression for the given value of the variables.

Rudik [331]

Since c=15 and d=3, we replace the c and d variables with 15 and 3.

0.5x15-1.7x3

0.5x15=7.5

1.7x3=5.1

Now subtract.

7.5-5.1=2.4

---

hope it helps

6 0

3 years ago

Find the gradients of line a and b

gayaneshka [121]

Answer:

Gradient of A: 2

Gradient of B: -1

Step-by-step explanation:

Gradient = change in y/change in x

✔️Gradient of A using two points on line A, (2, 5) and (0, 1):

Gradient = (1 - 5)/(0 - 2) = -4/-2

Simplify

Gradient of A = 2

✔️Gradient of B using two points on line B, (0, 5) and (5, 0):

Gradient = (0 - 5)/(5 - 0) = -5/5

Simplify

Gradient of B = -1

5 0

2 years ago

Sample annual salaries (in thousands of dollars) for employees at a company are listed. 42 36 48 51 39 39 42 36 48 33 39 42 45 (

Sonja [21]

Answer:

a) Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

$\bar X = 41.538$

$s = 5.317$

b) 44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

$\bar X = 43.615$

$s= 5.583$

c) 3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

$\bar X= 3.462$

$s = 0.443$

d) As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3.462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.

Step-by-step explanation:

For this case we can use the following formulas for the mean and standard deviation:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Part a

Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

And if we calculate the mean we got:

$\bar X = 41.538$

$s = 5.317$

Part b

For this case we know that each value present a 5% of rise so we just need to multiply each value bu 1.05 and we have this new dataset:

44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

And if we calculate the new mean and deviation we got:

$\bar X = 43.615$

$s= 5.583$

Part c

The new dataset would be each value divided by 12 so we have:

3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

And the new mean and deviation would be:

$\bar X= 3.462$

$s = 0.443$

Part d

As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3,462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.

5 0

3 years ago

A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is

tatiyna

Answer:

Hence to get same number of students in each classroom,the sufficient condition is that assign 13n students to each classroom.

Step-by-step explanation:

Given:

There are m classrooms and n be the students

3<m<13<n.

To Find:

Whether it is possible to assign each of n students to one of m classrooms with same no.of students.

Solution:

This problem is related to p/q form has to be integer in order to get same no of students assigned to the classroom.

As similar as ,n/m ratio

So 1st condition is that,

If it is possible to assign the n/m must be integer and n should be multiple of m,

when we assign 3n students to m classrooms ,we cannot say that 3n/m= integer so that n is greater than 13 i.e n=14 and m=6

hence they are not multiple of each other so they will not make same students in each classrooms.

Otherwise,n=14 and m=7 they will give same number but this condition is not sufficient condition to assign the student.

So 2nd condition is that ,

When we assign 13n students to m classrooms, as 13 is prime number and

3<m<13 which implies the 13n/m to be integer so n and m must be multiple of each other.

Suppose n=20 and m=5 classrooms

then 13*20=260 ,

260/5=52 students in each classroom,

4 0

3 years ago