3. f(-4) - 3.9(-2) =

After 1 minute, a submarine had descended to −280 feet. After 9 minutes, the submarine had descended to −360 feet. Assuming a li

adelina 88 [10]

Answer:

the answer is 640

Step-by-step explanation:

-280-360 = 640

d(t)=-360

= 360*d*t

4 0

3 years ago

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

What is a comparison of numbers or variables that can be written in fraction form

frozen [14]

I believe
4 and 6
4/1 and 6/1<span />

3 0

3 years ago

ILL GIVE BRAINELST

Angelina_Jolie [31]

Answer: number 3 4 5

Step-by-step explanation:hope this helps!

5 0

3 years ago

A manufacturer produces a commodity where the length of the commodity has approximately normal distribution with a mean of 6.5 i

zavuch27 [327]

The probablity that the sample's mean length is greate than 6.3 inches is0.8446.

Given mean of 6.5 inches,standard deviation of 0.5 inches and sample size of 46.

We have to calculate the probability that the sample's mean length is greater than 6.3 inches is 0.8446.

Probability is the likeliness of happening an event. It lies between 0 and 1.

Probability is the number of items divided by the total number of items.

We have to use z statistic in this question because the sample size is greater than 30.

μ=6.5

σ=0.5

n=46

z=X-μ/σ

where μ is mean and

σ is standard deviation.

First we have to find the p value from 6.3 to 6.5 and then we have to add 0.5 to it to find the required probability.

z=6.3-6.5/0.5

=-0.2/0.5

=-0.4

p value from z table is 0.3446

Probability that the mean length is greater than 6.3inches is 0.3446+0.5=0.8446.

Hence the probability that the mean length is greater than 6.3 inches is 0.8446.

Learn more about probability at brainly.com/question/24756209

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5 0

2 years ago