Someone help please

Hurry only 5 minutes to answer!!!!!!!!!!!!!1111

xz_007 [3.2K]

Answer:

-12, -3/4, -0.4, 0.4, 3/4, 8

Step-by-step explanation:

3 0

2 years ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

3 0

2 years ago

If you want to be 95% confident of estimating the population mean to within a sampling error of plus or minus3 and the standard

Aleksandr [31]

Answer:

The sample size required is at least 171

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What sample size is required?

A samle size of at least n is required, in which n is found when $M = 3, \sigma = 20$

So

$M = z*\frac{\sigma}{\sqrt{n}}$

$3 = 1.96*\frac{20}{\sqrt{n}}$

$3\sqrt{n} = 20*1.96$

$\sqrt{n} = \frac{20*1.96}{3}$

$(\sqrt{n})^{2} = (\frac{20*1.96}{3})^{2}$

$n = 170.7$

Rouding up

The sample size required is at least 171

5 0

2 years ago

2x+4y=16 what is the slope intercept form

Eva8 [605]

Since slope intercept is y=mx+b, you first want to get rid of the 2x by subtracting it from both sides. after, you should end up with 4y=16-2x or 4y=-2x+16, they both are the same, but some teachers might prefer the slope first. after that, you want to get the y alone, so you divide everything by 4. you should end up with the slope intercept form being y=-2/4x+4, simplified to y=-1/2x+4.

3 0

3 years ago

Estimate the product<br> A)Close to 40<br> B)Close to 004<br> C)Close to 0.4<br> D)Close to 4

PtichkaEL [24]

Close to 0.4 aka c :))

7 0

2 years ago

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