Which rational number equals 0.333...? 1 over 5, 1 over 3, 3 over 10, 3 over 5

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

2 years ago

What is the vertex of the Graph<br>MATH 1 PLEASE HELP

Pie

I can't see the problem so repost it and tag me in it

7 0

3 years ago

Greg runs 3 miles in 28 minutes. At the same rate, how many miles would he run in 42 minutes?

Paladinen [302]

Answer:

4.5

miles

Step-by-step explanation:

Well, he runs

3

miles in

28

minutes. Also, see that

42

=

28

⋅

1.5

.

So, his journey would just be running

1.5

times, with each time running

28

minutes.

So, he runs a total of

1.5

⋅

3

=

4.5

miles

7 0

2 years ago

Read 2 more answers

Suppose that the position of one particle at time is given by x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π and the position of a second p

Mashcka [7]

Answer:

there is no collision between the particles

Step-by-step explanation:

for the first particle

x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π

for the second particle

x2 = -3 + cos t, y2 = 1 + sin t, 0 ≤ t ≤ 2π

then for the collision

x1=x2 → 3*sin t = -3 + cos t → sin t= -1 + (cos t)/3→ 1+ sin t = (1/3)cos t

y1=y2 → 1 + sin t = 2 cos t → (1/3)cos t = 2 cos t →(1/3) = 2

since 1/3 ≠ 2 there is no collision between the particles

6 0

3 years ago

If 3 p +3 4 =90

klemol [59]

3p + 34 = 90
3p = 90 - 34
3p = 56
p = 56/3
p = 18.67 (thats rounded)

2r + 44 = 76
2r = 76 - 44
2r = 32
r = 32/2
r = 16

53 + 6s = 1421
6s = 1421 - 53
6s = 1368
s = 1368/6
s = 228

" the product " means multiply
p x r x s = 18.67 x 16 x 228 = 68,108.16

8 0

3 years ago