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yuradex [85]
4 years ago
7

Order least to greatest, 0.3, 19 over 40 ot 22%

Mathematics
2 answers:
Ksenya-84 [330]4 years ago
8 0
The answer is 0.3 , 19 , 22% , 40
emmainna [20.7K]4 years ago
6 0

Answer:

22%, 0.3, \frac{19}{40}

Step-by-step explanation:

To make this more simple to figure out, let's change everything into a percentage.

To change 0.3 to a percentage, simply move the decimal point right two spaces. Now you have 30%.

Now to change \frac{19}{40} to a percentage. Simply divide 19 by 40.

19 ÷ 40 = 0.475

Now we do what we did above; change the decimal to a percentage.

The decimal is now 47.5%.

22% is already a percentage.

Now, let's compare.

30%, 47.5%, 22%

The greatest number seems to be 47.5%, while the least is 22%.

Let's organize it from least to greatest.

22%, 30%, 47.5%

Now, change the percentages back to their original numbers.

47.5% = \frac{19}{40}

30% = 0.3

22% = 22%

22%, 0.3, \frac{19}{40}

I hope this helps! Please let me know if you anything else. :)

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guajiro [1.7K]
<span>The answer should be 4.5 because they used AT LEAST 32 miles per gallon. So, 144/32=4.5 and since it's AT LEAST 32, 4.5 is the max amount used.</span>
5 0
3 years ago
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Evaluate 1/4c+3d when c=6 and d=7
labwork [276]

\dfrac{1}{4}c+3d

Put the values of c = 6 and d = 7 to the expression:

\dfrac{1}{4}\cdot6+3\cdot7=\dfrac{1}{2}\cdot3+21=1.5+21=22.5

5 0
4 years ago
a triangle has an area of 54 m2 and a height of 9m how long is the base of the triangle entered your answer in the box
Delvig [45]
Answer:
base of the triangle = 12 m

Explanation:
Area of the triangle = 1/2 * base * height
We are given that:
area = 54 m^2
height = 9 m

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Area of the triangle = 1/2 * base * height
54 = 1/2 * base * 9
108 = 9*base
base = 12 m

Hope this helps :)
7 0
3 years ago
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A 250 mL beaker can hold 0.192 kg of acetone. What is the density of this substance, in g/mL? Round your answer to 3 significant
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3 0
4 years ago
A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
AURORKA [14]

Answer:

(−0.103371 ; 0.063371) ;

No ;

( -0.0463642, 0.0063642)

Step-by-step explanation:

Shift 1:

Sample size, n1 = 30

Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm

Shift 2:

Sample size, n2 = 25

Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17

Mean difference ; μ1 - μ2

Zcritical at 95% confidence interval = 1.96

Using the relation :

(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)

(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)

Lower boundary :

-0.02 - 0.0833710 = −0.103371

Upper boundary :

-0.02 + 0.0833710 = 0.063371

(−0.103371 ; 0.063371)

B.)

We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.

C.)

For sample size :

n1 = 300 ; n2 = 250

(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)

Lower boundary :

-0.02 - 0.0263642 = −0.0463642

Upper boundary :

-0.02 + 0.0263642 = 0.0063642

( -0.0463642, 0.0063642)

7 0
3 years ago
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