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finlep [7]
3 years ago
5

Question in the attachment!!!

Mathematics
1 answer:
IgorC [24]3 years ago
6 0
Reflection along x axis
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If the first term is 3 (A1=3) and the common difference is 4 (d=4) what is the fifth term? (A5=?)​
omeli [17]

Answer:

Pov:شما به ترجمه گوگل رفت برای دیدن آنچه من گفتم و دیدم این xD

Step-by-step explanation:

Pov:شما به ترجمه گوگل رفت برای دیدن آنچه من گفتم و دیدم این xD

3 0
3 years ago
Find the distance between the points (11, 4) and (10, 5). 1 2
Debora [2.8K]
Distance formula : d = sqrt (x2 - x1)^2 + (y2 - y1)^2
(11,4)(10,5)

d = sqrt (10 - 11)^2 + (5 - 4)^2
d = sqrt (-1^2) + (1^2)
d = sqrt (1 + 1)
d = sqrt 2
d = 1.41 <===
3 0
4 years ago
Read 2 more answers
5/6 + 7/8 + 3/4 <br> i need help
Dimas [21]

Answer:2 11/24

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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
Write the algebraic expression for each word problem. See if you can spot the trick problem that doesn’t need algebra!
AfilCa [17]

Answer:

6) -7 + 10 = 3 (3 more; total number doesn't matter)

7) 50w + 20

8) c / 5

9) 5 + 2t

10) 4d  + 2d

I didn't see the trick problem!

7 0
3 years ago
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