What is 81.139 rounded to the nearest tenth

You practice playing your guitar every day. You spend 15 minutes practicing chords and the rest of the time practicing a new son

Black_prince [1.1K]

Answer:

Step-by-step explanation:

Given that the total time taken to practice is given by the expression as

y=7(15+x)

Simplifying the expression we have

y=105+7x

Solving for x (that is making x subject of formula we have)

7x=y-105

Divide both sides by 7 we have

x=(y-105)/7

Therefore the expression is

x=(y-105)/7

5 0

3 years ago

...........answers plz

Ira Lisetskai [31]

Answer:

0

Step-by-step explanation:

m = 21, n = 12

so we just plug these into the equation

(1/3)(21) -1 -(1/2)(12)

(21/3) -1 -(12/2)

7 -1 -6

6 -6

0

3 0

3 years ago

Help!!! 30pts for whoever answers!! Please helpppp.

VARVARA [1.3K]

Answer:

i am srry i dont know but maybe u can multiply the cost per day by cost per mile then subtract or divide

by 2

4 0

3 years ago

Read 2 more answers

Students attending the University of Florida can select from 130 major areas of study. A student's major is identified in the re

omeli [17]

Answer:

1. 8385 double majors

2. 18252 major codes

3. 8515 codes

4. Yes

Step-by-step explanation:

Given

Major areas = 130

Alphabets = 26

a.

There are 130 major areas in the university.

To get a double majors, means one select any of two of the major areas in the university.

i.e. just 2 alphabets are needed

Number of Double Majors = 130C2

130C2 = 130!/(128!*2!)

130C2 = 130 * 129/2

130C2 = 8385.

b.

Number of major codes available = Number of 2 digit codes + Number of 3 digit codes

Number of 2 digit codes = 2 alphabets

There are 26 ways of selecting the first letter

There are 26 ways of selecting the second letter

So, number of 2 digit codes = 26 * 26 = 676.

Number of 3 digit codes = 3 alphabets

There are 26 ways of selecting the first letter

There are 26 ways of selecting the second letter

There are 26 ways of selecting the third letter

So, number of 3 digit codes = 26 * 26 * 26 = 17576

Thus, number of major codes = 17576 + 676 = 18252

c.

Number of codes to identify a student with either a single major or double major

Number off single major codes = number of major areas = 130

Number of double major codes = 8385

Total = 130 + 8385 = 8515

d. From the solutions above, Yes there are enough codes available to identify all single and double majors at the university

6 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

Alexxandr [17]

Answer:

P = 0,0012 or P = 0.12 %

Step-by-step explanation:

We know for normal distribution that:

μ ± σ in that range we find 68.3 % of all values

μ ± 2σ ⇒ 95.5 % and

μ ± 3σ ⇒ 99.7 %

Fom problem statement

We have to find (approximately) % of cars that reamain in service between 71 and 83 months

65 + 6 = 71 ( μ + σ ) therefore 95.5 % of values are from 59 and up to 71 then by symmetry 95.5/2 = 49.75 of values will be above mean

Probability between 65 and 71 is 49.75 %

On the other hand 74 is a value for mean plus 1, 5 σ and

74 is the value limit for mean plus 1,5 σ and correspond to 49,85 (from z=0 or mean 65).

Then the pobabilty for 83 have to be bigger than 49.85 and smaller than 0,5 assume is 49.87

Finally the probability approximately for cars that remain in service between 71 and 83 months is : 0,4987 - 0.49.75

P = 0,0012 or P = 0.12 %

8 0

3 years ago