All categories

3 years ago

Is this rational or irrational? Then explain why is this rational or irrational

Mathematics

2 answers:

irga5000 [103]3 years ago

6 0

Answer:

Irrational

Step-by-step explanation:

.1212 repeated is irrational because 1, it cannot be written as a fraction and 2, it will never end.

xxMikexx [17]3 years ago

4 0

Answer:

It is Irrational

Step-by-step explanation:

Rational: it can be expressed as an fractions and it can stop and repeat

Irrational: goes on forever when in decimal form and the numbers don't repeat

You might be interested in

What is another way to write (3x1000)+(3x10)+(3x1/100)+(3x1/1000

Paha777 [63]

That'd be 3000 + 30 + 0.03 + 0.003, or 3030.033

hope this helped :)
alisa202

plzz tell me if I was wrong

3 0

3 years ago

Find the prime factorization of the number 8.

anyanavicka [17]

Answer:

For instance, the prime factorization of 8 is 2×2×2, not just "2". Yes, 2 is the only factor, but you need three copies of it to multiply back to 8, so the prime factorization includes all three copies.

Step-by-step explanation:

Very intresting hope this helps

7 0

3 years ago

An antique car cost $32,00. It gains 8% in value every year. How much will it be worth in 6 years

denis23 [38]

If you mean $3,200 the answer is $5088

If you meant $32,000 the answer is $50779.98

6 0

3 years ago

Help is it A or B? pls help

Damm [24]

Answer:

$f(x)=-(x+8)(x-2)$

Step-by-step explanation:

Start by factoring out a -1...

$f(x)=-(x^2+6x-16)$

Now, we have to find two integers that multiply to get -16 and have a sum of 6:

(-2)*8=-16

-2+8=8-2=6

Using this, we can split 6x into -2x and 8x...

$f(x)=-(x^2-2x+8x+16)$

Factor the first and second half separately...

$f(x)=-[(x^2-2x)+(8x-16)]\\f(x)=-[x(x-2)+8(x-2)]\\$

Since both x and 8 are being multiplied by x-2, we can combine them to get...

$f(x)=-(x+8)(x-2)$

5 0

2 years ago

4. The average annual income of 100 randomly chosen residents of Santa Cruz is $30,755 with a standard deviation of $20,450. a)

Umnica [9.8K]

Answer:

a) The standard deviation of the annual income σₓ = 2045

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to $33,000

d)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

Step-by-step explanation:

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation = $20,450

The standard deviation of the annual income σₓ = $\frac{S.D}{\sqrt{n} }$

= $\frac{20,450}{\sqrt{100} }= 2045$

Given mean of the Population μ = $32,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ > $32,000

Alternative Hypothesis:H₁: μ < $32,000

Level of significance α = 0.10

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-32000 }{\frac{20450}{\sqrt{100} } }$

Z= |-0.608| = 0.608

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

Given mean of the Population μ = $33,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ = $33,000

Alternative Hypothesis:H₁: μ ≠ $33,000

Level of significance α = 0.05

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-33000 }{\frac{20450}{\sqrt{100} } }$

Z = -1.0977

|Z|= |-1.0977| = 1.0977

The 95% of z -value = 1.96

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to $33,000

d)

95% of confidence intervals is determined by

$(x^{-} - 1.96 \frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} })$

$(30755 - 1.96 \frac{20450}{\sqrt{100} } , 30755 +1.96 \frac{20450}{\sqrt{100} })$

( 30 755 - 4008.2 , 30 755 +4008.2)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

4 0

3 years ago