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kondor19780726 [428]

2 years ago

6

Select the system of linear inequalities whose solution is graphed.

1 answer:

Volgvan2 years ago

8 0

The second one is your answer

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At a large Midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen

guajiro [1.7K]

Answer:

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

p1 -> 1993

20 out of 100, so:

$p_1 = \frac{20}{100} = 0.2$

$s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

p2 -> 1997

10 out of 100, so:

$p_2 = \frac{10}{100} = 0.1$

$s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

Distribution of p1 – p2:

$p = p_1 - p_2 = 0.2 - 0.1 = 0.1$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05$

Confidence interval:

$p \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower bound of the interval is:

$p - zs = 0.1 - 1.645*0.05 = 0.01775
$

The upper bound of the interval is:

$p + zs = 0.1 + 1.645*0.05 = 0.18225
$

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

6 0

3 years ago

Please help me with 1-4 and 6 and 7

Tom [10]

I could only get number one which was; 1.75

3 0

3 years ago

Can someone pls help

MakcuM [25]

when x =1 , y = 3

when x = 3, y =27

27-3 = 24

3-1 = 2

24/2 = 12

rate of change = 12

3 0

3 years ago

The ratio of the number of left-handed batters to the number of right handed batters is 5:8. There are 45 left-handed batters. H

OleMash [197]

5/8=45/x
45×8=5x
360=5x
360÷5
72

3 0

2 years ago

Find the volume of the following solid. The solid between the cylinder f(x,y)equals=e Superscript negative xe−x and the region

PilotLPTM [1.2K]

It is hard to comprehend your question. As far as I understand:

f(x,y) = e^(-x)

Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.

That is all I understood. It would be easier to understand with a picture or some kind of visual aid.

Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy$

Now evaluate,

$\int_{0}^{ln(6)}e^{-x}dx$

which evaluates to, 5/6 if I did the math correct. Correct me if I am wrong.

Now integrate this w.r.t. y:

$\int_{-6}^{6}\frac{5}{6}dy = 10$

So,

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10$

7 0

3 years ago