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3 years ago

How do you find the mean, median and range

Mathematics

2 answers:

ivolga24 [154]3 years ago

6 0

Answer:

Mean= Average

Median= Middle number after numbers are lined up in order

Mode= Repeated Number

Step-by-step explanation:

astra-53 [7]3 years ago

5 0

for mean you add every number and divide it by the amount of numbers in the data set

for median you put the numbers in order and find the middle number, if there is no middle get the mean for the two middle numbers

range is basically the lowest and highest number

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It takes Bruce 12 h to complete his training. It takes Oliver approximately 2 days, 2 h, and 15 min to complete

Stolb23 [73]

Answer:

Oliver completed his training work in 50.25 hours

Step-by-step explanation:

we know that

1 day= 24 hours

1 hour=60 minutes

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2 days, 2 h, and 15 min

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4 years ago

Two shooters, Rodney and Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark

Sedaia [141]

Complete Question

Answer:

$SE = 0.66}$

$-3.29 < \mu_1 - \mu_2 < -0.70$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

$\sigma_1 = \sqrt{v_1}$

=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

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4 years ago

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nordsb [41]

Just like looking at a mirror image of yourself, but flipped....a reflection point is the mirror point on the opposite axis.

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4 years ago

How do you find the mean, median and range​

How do you find the mean, median and range