Suppose candidate a for a town council seat receives 43% of the votes in an election. as voters leave the polls they are asked who they voted for. what is the probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a? assume an infinite population.

The probability that less than 40% of the 80 voters surveyed indicate they voted for candidate A is P(phat <0.4) = P((phat-p)/sqrt(p*(1-p)/n) <(0.4-0.43)/sqrt(0.43*(1-0.43)/80)) =P(Z<-0.542) =0.2939 (from standard normal table)

29.46% probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For a proportion p in a sample of size n, we have

In this problem, we have that:

What is the probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a?

This is the pvalue of Z when X = 0.4. So

has a pvalue of 0.2946

29.46% probability that less than 40% of the 80 voters surveyed indicate they voted for candidate a

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

the car was traveling 40 mph before it started skidding. ( 2 points ) 17. was the driver were her brakes working at 75% efficiency or better? ( 2 points ) yes 29.

step-by-step explanation: