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3 years ago

10

A towns population increased from 14523 to 16489 what is the percent increase in the towns population

2 answers:

Alenkasestr [34]3 years ago

8 0

First you need to figure out how much it increased by. 16489-14523= 1966
The population increased by 1966. Now you just need the percentage. You should get the percentage by dividing the increase in population by the original population. That would be 1966/14523=0.135. To get the percentage you must multiply 0.135 by 100.
0.135 x 100= 13.5%
That should be the answer.

MrRissso [65]3 years ago

5 0

Answer:14%

Step-by-step explanation:

Your welcome

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A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len

Gnom [1K]

Answer:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

$df=n-1=93-1=92$

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

$\bar X=5.97$ represent the sample mean for the length

$s=0.4$ represent the sample standard deviation

$n=93$ sample size

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:

Null hypothesis: $\mu = 6$

Alternative hypothesis: $\mu \neq 6$

since we don't know the population deviation the statistic is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing in formula (1) we got:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

P value

The degrees of freedom are given by:

$df=n-1=93-1=92$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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