Using the Distance Formula
1 answer:
Answer:
Step-by-step explanation:
since I did 1 already , i'll do 2 - 6 .. fast
using the formula for distance = sq rt [ (x2 - x1 ) ^2 + (y2 - y1 ) ^2 ]
2) sq rt [ (6-1)^2 + (14-2)^2
sq rt [ 25 + 144]
sq rt [ 169]
13
3) sq rt [ (15 - 8)^2 + (35-11)^2 ]
sq rt [ 49 + 576 ]
sq rt [625]
25
4) sq rt [ ( -5 -3 )^2 + (-7 - 8)^2 ]
sq rt [ 64 + 225]
sq rt [ 289]
17
5) sq rt [(-1-(-2))^2 + (-4-(-3))^2]
sq rt[ (-1+2)^2 + (-4+3)^2]
sq rt [ 1 + 1 ]
6) sq rt [ 3-(-3)^2 + (5-5)^2]
sq rt [ (3+3)^2 + 0]
sq rt [ 9 ]
3
:) phew.... :P
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Answer:
Yes, this provide enough evidence to show a difference in the proportion of households that own a desktop.
Step-by-step explanation:
We are given that National data indicates that 35% of households own a desktop computer.
In a random sample of 570 households, 40% owned a desktop computer.
<em><u>Let p = population proportion of households who own a desktop computer</u></em>
SO, Null Hypothesis, : p = 25% {means that 35% of households own a desktop computer}
Alternate Hypothesis, : p
25% {means that % of households who own a desktop computer is different from 35%}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of 570 households who owned a desktop computer = 40%
n = sample of households = 570
So, <u><em>test statistics</em></u> =
= 2.437
<em>Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>
Therefore, we conclude that % of households who own a desktop computer is different from 35%.
Answer:
Step-by-step explanation:
