statzzz
@Michele_Laino
I was wondering if you could check my answers...
thanks!
question #4, I got \[\Large \left( {0.7242,\quad 0.8358} \right)\]
Solve negative 7 over 3, the whole multiplied by x minus 3 equals negative 52. 21 39 45 59
Need help
question #9 I got 2401, namely the last option. Please you have to reply, since I can not give you the answer directly
ok sorry
i just got back
and got it and thanks
are you on 10?
Help
question #10 my definition of error margin m, is: \[m = \frac{{\Delta a}}{a}\] where a is the mean value, and \Delta a is the uncertainty on a
how have you defined the error margin?
ok
Ok ill wait
@Michele_Laino are all the others right?
I'm pondering...
ok np
for example, in question #3 how do you got 745?
can you write your steps, in order to get 745?
it was so long ago when I did these i forgot the steps but remembered the ansers that is why im asking for u to check them lol
ok! do you remember the definition of error margin?
question #3 the standard deviation is: 11.534 so the 95% confidence interval is 11.534* 1.65 = 19
ok thnxs sorry for the late reply
The graph below shows the prices of different numbers of boots at a store: A graph is shown. The values on the x axis are 0, 2, 4, 6, 8, and 10. The values on the y axis are 0, 25, 50, 75, 100, and 125. Points are shown on ordered pairs 0, 0 and 2, 25 and 4, 50 and 6, 75 and 8, 100. These points are connected by a line. The label on the x axis is Number of Boots. The title on the y axis is Price in dollars. Which equation can be used to determine p, the cost of b boots? p = 12.50b p = 25b b = 25p b = 12.50p
question #5 I got 261, namely, close to the first option
question #3 I got n=751
nice and thnxs!
question #6 I think you are right! we have to determine the new proportion
question #10 hint: the formula for margin error, is: \[\Large m = {z^*}\sqrt {\frac{{p\left( {1 - p} \right)}}{n}} \] where p is the sample proportion, n is the sample size, and z* is the value which comes from the percentage of the confidence level
question #7 I got: \[\Large \left( {0.24316,\quad 0.35684} \right)\]
the procedure is the same as in the first exercise, we have to apply the binomial distribution
question #8, I got: n= 504.21, so your answer is right!
awesome! thnxs
finally, for question #9, I have applied the subsequent formula: \[\Large \frac{{1.96}}{{\sqrt n }} = 0.04\]
our homework is completed!
yah!!1 can u help later? if so, can u give the hours your online
I will stay here for at least 2 hours
ok cool can u help some more then plz?
ok!
one second plz, give me 10 minutes before i upload some more and ty soo much
@Michele_Laino , is #10 right?
question #10 I give this hint: \[m = {z^*}\sqrt {\frac{{p\left( {1 - p} \right)}}{n}} \] m is the margin of error, p is the sample proportion, n is the sample size, and z* is coming from the percentage of confidence level, namely if confidence level is 95%, then z*=1.96, if confidence level is 90%, then z*=1.65
I think option 3 only
ok
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