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shepuryov [24]
3 years ago
5

Off Topic: when you delete a place on Life360, does the other person in your circle get the notification that you have remove it

?yes or no?
Mathematics
1 answer:
nlexa [21]3 years ago
8 0

Answer:

don't think so

Step-by-step explanation:

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Round 17 5/8 to the nearest whole number
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An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand
Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

b) We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

Step-by-step explanation:

Previous concepts

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

4 0
3 years ago
A treatment is administered to a sample of n = 9 individuals selected from a population with a mean of µ = 80 and a standard dev
stepan [7]

Answer:

a) M =82

Step-by-step explanation:

Let´s study this as a Normal distribution.

As we know in a normal distribution the z score is = (X-μ)/(sd/sqrt(n))

where

X = mean for the taken sample = What we want to know in this problem

μ = Total population mean =80

sd= standard deviation= 12

n = sample size=9

So in this case z=( X-80)/(12/sqr(9))= (X-80)/4

also we know that the effect size taken by the machine is 0.5, which is the same as the z-score

so...

0.5 = (X-80)/4 => 0.5*4 = X-80

2+80 = X

X = 82

4 0
3 years ago
A market sold two packages of sliced beef weighing 1 and one seventh and 4 and one half lb. What was the total weight of the bee
Pepsi [2]

Answer:

5 and 9/14

Step-by-step explanation:

convert the numbers into improper fractions

8/7 + 9/2

=

(8 × 2) + (9 × 7)

7 × 2

=

79/14   or 5 and 9/14


7 0
3 years ago
Read 2 more answers
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