All categories

3 years ago

How many times does 81,000 goes into 8,000

Mathematics

1 answer:

Effectus [21]3 years ago

8 0

Zero. That number cannot go into that one

You might be interested in

Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr

Minchanka [31]

Answer:

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]p_1$ represent the real population proportion for 1

$\hat p_1 =0.768$ represent the estimated proportion for 1

$n_1=92$ is the sample size required for 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.646$ represent the estimated proportion for 2

$n_2=95$ is the sample size required for 2

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

$ME= \frac{0.2753-(-0.0313)}{2}=0.1533$

And we know that the margin of erro is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

We have all the values except the value for $z_{\alpha/2}$

So we can find it like this:

$0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}$

And solving for $z_{\alpha/2}$ we got:

$z_{\alpha/2}=2.326$

And we can find the value for $\alpha/2$ with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98$ or 98%

7 0

3 years ago

41 out of 50 women surveyed liked perfume. what percentage of women did NOT like perfume

krok68 [10]

Answer:

18%

Step-by-step explanation:

Long Version: 41/50 = 0.82 and when you move the decimal two places right, you will get 82%. 100%-82%= 18%

Short Version: 9/50 people didn't like perfume. 9/50= 0.18 then you move the decimal two places right and it becomes 18%.

3 0

3 years ago

Read 2 more answers

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

The complex numbers corresponding to the endpoints of one diagonal of a square drawn on a complex plane are 1 + 2i and -2 – i.

kodGreya [7K]

Technically the endpoints will be intersection of first endpoint between x=1 and y=-1second endpoint between x=-2 and y=2
so they are 1-i and -2+2i

7 0

3 years ago

Help pls do allll 6 questions... with steps

BaLLatris [955]

You just want to simplify right?!

45. (a^2b^3)(ab)^-2
= (a^2b^3)(a^-2b^-2)
= b
46. (-3x^3y)^2(4xy^2)
= (-9x^6y^2)(4xy^2)
= -36x^7y^4
47. 3c^2d(2c^3d^5) / 15c^4d^2
= 6c^5d^6 / 15c^4d^2
= 2/5c1/4x^4
48. -10g^6h^9(g^2h^3) / 30g^3h^3
= -10g^8h^12 / 30g^3h^3
= -1/3g^5h^9
49. 5x^4y^2(2x^5y^6) / 20x^3y^5
= 10x^9y^8 / 20x^3y^5
= 1/2x^6 1/3y^3
50. -12n^7p^5(n^2p^4) / 36n^6p^7
= -12n^9p^9 / 36n^6p^7
= -1/3n^3p^2

(Sorry it’s messy it’d look better if my phone could actually put the numbers to the power)

5 0

3 years ago