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2 years ago

State the excluded value of 60x^3 /12x

Mathematics

1 answer:

Neporo4naja [7]2 years ago

8 0

Answer:

$\frac{60 {x}^{3} }{12x}$

$5 {x}^{2}$

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Sorry if im asking too much im not the best with math/ it's not my best subject

Naya [18.7K]

Find the mean:

6 + 6 +10 + 8 + 10 +8 = 48

48 / 6 = 8

The mean is 8.

Now for each number, subtract the mean and square the result:

6 -8 = -2, -2^2 = 4

6 – 8 = -2, -2^2 = 4

10 – 8 = 2, 2^2 = 4

8 – 8 = 0, 0^2 = 0

10 – 8 = 2, 2^2 = 4

8 – 8 = 0, 0^2 = 0

Now find the mean of those results:

4 + 4 +4 +0 +4 + 0 = 16

16 / 6 = 2.6666

Now find the square root:

SQRT(2.666) = 1.63

The answer is A.

6 0

2 years ago

Question 1 plz show ALL STEPS

abruzzese [7]

Answer:

1a= 10

Step-by-step explanation:

$f(g(1) = 3( - 2(1) + 7) - 5$

$= 3(5) - 5$

$15 - 5 = 10$

7 0

2 years ago

Read 2 more answers

N–52=12 answer plssss

chubhunter [2.5K]

n-52=21

n+52=+52

N=72

dk cnfn evf djokfwdvi jdvofjoirvuowe0pwv

6 0

2 years ago

Read 2 more answers

Johan's family took him out to dinner to celebrate his amazing 2nd

leonid [27]

Answer:

$190.32

Step-by-step explanation:

$175.00 x 1.08756=$190.32

(1.08756 represents 1 + 8.756%)

8 0

2 years ago

According to records, the amount of precipitation in a certain city on a November day has a mean of 0.10 inches, with a standard

Andru [333]

Answer:

The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days

P(X≤ 0.11) = 0.4404

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 0.10 inches

Given that the standard deviation of the population = 0.07inches

Let 'X' be a random variable in a normal distribution

$Z = \frac{x-mean}{S.D} = \frac{0.11-0.10}{0.07} = 0.1428$

Step(ii):-

The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days

P(X≤ 0.11) = P(Z≤0.1428)

= 1-P(Z≥0.1428)

= 1 - ( 0.5 +A(0.1428)

= 0.5 - A(0.1428)

= 0.5 -0.0596

= 0.4404

Final answer:-

The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days

P(X≤ 0.11) = 0.4404

8 0

2 years ago