1. An ant starts from the origin of a 2-D grid, at the point (0,0). Then each time it moves either i) right by one unit distance, e. g., from (0,0) to (1,0), ii) up by one unit distance, e. g., from (0,0) to (0,1), or iii) stay still, with probabilities 0.3, 0.2, and 0.5, respectively. After 10 moves, what is the probability that the ant ends up at point (2,5)

0.0091 = 0.91% probability that the ant ends up at point (2,5)

Step-by-step explanation:

What is needed to move from the point (0,0) to the point (2,5) after 10 moves:

5 moves of up by 1 unit.

2 moves of right by 1 unit.

10 - (5 + 2) = 3 standing still.

Probability:

2 moves of right by 1 unit, each with 0.3 probability.

5 moves of up by 1 unit, each with 0.2 probability.

3 standing still, each with 0.5 probability.

Number of ways this can happen:

Arrangments of 10 moves, with 2, 5 and 3 repetitions. So

Probability:

0.0091 = 0.91% probability that the ant ends up at point (2,5)

domain is all the x values included in the

0 ≤ x ≤ 2

range is all the y values of the graph

-2 ≤ y ≤ 2

***keep in mind that there are different ways to show domain and range

hope this !

~just a girl in love with shawn mendes

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;