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BartSMP [9]
3 years ago
10

Is integer number divided by integer number closed or not closed

Mathematics
2 answers:
Semenov [28]3 years ago
7 0

Answer:

Integers are Not Closed under Division.

Step-by-step explanation:

There are three integers between-2 and 2. ... The set of integers is closed under addition, subtraction, and multiplication because when I add, subtract, or multiply any integers, the result is always an integer. The set is not closed under division because 2 + 4 = 0.5, and 0.5 is not an integer. The integers are "closed" under addition, multiplication and subtraction, but NOT under division ( 9 ÷ 2 = 4½). (a fraction) between two integers. Integers are rational numbers since 5 can be written as the fraction 5/1.

iren2701 [21]3 years ago
5 0

Answer:

Integers are Not Closed under Division

Step-by-step explanation:

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Y = –3x – 5 help I don’t know the answer please
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Answer:

x=-5/3

Step-by-step explanation:

Add 5 to both sides to get 5=-3x.

Then divide by -3.

And get -5/3=x

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Diego’s family car holds 14 gallons of fuel. Each day the car uses 0.5 gallons of fuel. A warning light comes on when the remain
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His father can drive 25 days without the warning light
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Find a polynomial, which, when added to the polynomial 5x^2–3x–9, is equivalent to <br><br> x^2-5x+6
alexandr1967 [171]
So this is easy counting
so 5x^2-3x-9
we have 5 x^2's
-3 x's
-9

wat do we add to that to get
1 x^2's
-5 x's
+6
so we do each
5x^2+what=1x^2
subtract 5x^2
from both sides
what=-4x^2
first is -4x^2

then we have -3x to -5x
-3x+what=-5x
add 3x to both sides
what=-2x
second is -2x

-9 to 6
-9+ what=6
ad d9
what=15
last is 15
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6 0
3 years ago
Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra
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<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n= sample size

s= sample standard deviation

t_{\alpha/2} = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , \overline{x}=22.4

\alpha=1-0.99=0.01

A) Critical t value for \alpha/2=0.005 and degree of 18 freedom

t_{\alpha/2} = 2.8784

B) Required confidence interval:

22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

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