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34kurt
3 years ago
8

Six times a number plus 4 is the same as the numver minus 11

Mathematics
1 answer:
Brrunno [24]3 years ago
3 0
6xn+n-11 so n=6

6x6=36+6=42-11 is 31
You might be interested in
Three multiplied by the sum of 4 and a number is the same as 12 more than the number. Find the number.
Bess [88]

Answer:

3(4+n) = 12+n is an equation that could be used.

12n = 12+n

11n = 12

n= 1+1/11

Step-by-step explanation:

7 0
2 years ago
A department within an agency is randomly chosen each week for auditing. If there are 25 departments, what is the chance that an
Alexandra [31]

<u>Answer:</u>

The chance that any single department is chosen for auditing in a given week is 0.04%

<u>Explanation:</u>

Given that a department within an agency is randomly chosen per week for auditing and we are given total number of departments are 25 departments of which 1 is to be chosen.

Now chance of any single department to be chosen in any particular week will be given by 1 dividing by total no. of departments

which is equal to 1/25

1/25 = 0.04%

which is the chance of percentage of choosing a department

8 0
3 years ago
3/5 x 1/5+1/5 x 73/3
insens350 [35]

So,

3/5 x 1/5 = 3/25

3/25 + 1/5 = 8/25

8/25 x 73/3 = 7 59/75

I hope this helps

5 0
3 years ago
Why are there 27 cubic feet in one cubic yard
kobusy [5.1K]

One cubic yard is equal to 27 cubic feet because one yard is equal to 3 feet

<em><u>Solution:</u></em>

Given that, there 27 cubic feet in one cubic yard

Yard and feet are units of distance

1 Yard (yd) is equal to 3 feet (ft). To convert yards to feet, multiply the yard value by 3.

<em><u>The conversion factor for 1 yard to 1 feet is given as:</u></em>

1 \text{ yard } = 3 \text{ foot }

Now find for 1 cubic yard . Cubic yard means yard raised to power 3

Therefore,

1\ yard^3 = (3)^3\ feet^3\\\\1\ yard^3 = 27 feet^3\\\\1\ cubic\ yard = 27\ cubic\ feet

Thus one cubic yard is equal to 27 cubic feet

Therefore, there are 27 cubic feet in one cubic feet

8 0
2 years ago
An example of an early application of statistics was in the year 1817. A study of chest circumference among a group of Scottish
otez555 [7]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: chest circumference of a Scottish man.

X≈N(μ;δ²)

μ= 40 inches

δ= 2 inches

The empirical rule states that

68% of the distribution lies within one standard deviation of the mean: μ±δ= 0.68

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

99% of the distribution lies within 3 standard deviations of the mean: μ±3δ= 0.99

a)

The 58% that falls closest to the mean can also be referred to as the middle 58% of the distribution, assuming that both values are equally distant from the mean.

P(a≤X≤b)= 0.58

If 1-α= 0.58, then the remaining proportion α= 0.42 is divided in two equal tails α/2= 0.21.

The accumulated proportion until "a" is 0.21 and the accumulated proportion until "b" is 0.21 + 0.58= 0.79 (See attachment)

P(X≤a)= 0.21

P(X≤b)= 0.79

Using the standard normal distribution, you can find the corresponding values for the accumulated probabilities, then using the information of the original distribution:

P(Z≤zᵃ)= 0.21

zᵃ= -0.806

P(Z≤zᵇ)= 0.79

zᵇ= 0.806

Using the standard normal distribution Z= (X-μ)/δ you "transform" the values of Z to values of chest circumference (X):

zᵃ= (a-μ)/δ

zᵃ*δ= a-μ

a= (zᵃ*δ)+μ

a= (-0.806*2)+40= 38.388

and

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (0.806*2)+40= 41.612

58% of the chest measurements will be within 38.388 and 41.612 inches.

b)

The measurements of the 2.5% men with the smallest chest measurements, can also be interpreted as the "bottom" 2.5% of the distribution, the value that separates the bottom 2.5% of the distribution from the 97.5%, symbolically:

P(X≤b)= 0.025 (See attachment)

Now you have to look under the standard normal distribution the value of z that accumulates 0.025 of the distribution:

P(Z≤zᵇ)= 0.025

zᵇ= -1.960

Now you reverse the standardization to find the value of chest circumference:

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (-1.960*2)+40= 36.08

The chest measurement of the 2.5% smallest chest measurements is 36.08 inches.

c)

Using the empirical rule:

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

(μ-2δ) ≤ Xc ≤ (μ+2δ)=0.95 ⇒ (40-4) ≤ Xc ≤ (40+4)= 0.95 ⇒ 36 ≤ Xc ≤ 44= 0.95

d)

The measurements of the 16% of the men with the largest chests in the population or the "top" 16% of the distribution:

P(X≥d)= 0.16

P(X≤d)= 1 - 0.16

P(X≤d)= 0.84

First, you look for the value that accumulates 0.84 of probability under the standard normal distribution:

P(Z≤zd)= 0.84

zd= 0.994

Now you reverse the standardization to find the value of chest circumference:

zd= (d-μ)/δ

zd*δ= d-μ

d= (zd*δ)+μ

d= (0.994*2)+40= 41.988

The measurements of the 16% of the men with larges chess are at least 41.988 inches.

I hope this helps!

8 0
3 years ago
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