Observe the map of the United States above. Areas indicated in red are where tornadic activity is most likely. Which

Bill, George, and Ross, in order, roll a die. The first one to roll an even number wins and the game is ended. What is the proba

hram777 [196]

Answer: $0.5714\frac{1}{15}$

Step-by-step explanation:

Given that Bill, George, and Ross, in order, roll a die.

The first one to roll an even number wins and the game is ended.

Since Bill starts the game he can win by throwing even number or lose by throwing odd number

P(win) = 0.5, otherwise, the die will go to George. For Bill to win, both George and Ross should throw an odd number so that Bill again gets the chance with game non ending.

Thus we have Prob of Bill winning =P of Bill winning in I throw +P of Bill winning in his II chance of throw +....infinitely

To get back the dice once he loses probability

= p both throws odd = $0.5(0.5) = 0.25$

Thus Prob for Bill winning

= $0.5+0.25(0.5)^2+0.25(0.5)^3(0.25)+...$

This is an infinite geometric series with I term 0.5 and common ratio 0.125<1

Sum = $\frac{a}{1-r} =\frac{0.5}{1-0.125} =0.5714$

4 0

2 years ago

at what x coordinate would a line whose equation is y=2x-3 intersect a perpendicular line whose y intercept is 17

Jlenok [28]

Answer:

Step-by-step explanation:hj

4 0

1 year ago

Kelly ordered 10 pizzas for a party. 30% of the pizzas have 12 slices each. The remaining 70% of the pizzas have 8 slices each.

abruzzese [7]

Answer:

The total number of slices is 92

Step-by-step explanation:

we know that

Kelly ordered 10 pizzas for a party

step 1

30% of the pizzas have 12 slices each

so

The number of pizzas for 30% is equal to

0.30(10)=3 pizzas

The number of slices is

3(12)=36 slices

step 2

70% of the pizzas have 8 slices each

so

The number of pizzas for 70% is equal to

0.70(10)=7 pizzas

The number of slices is

7(8)=56 slices

step 3

Find the total number of slices

36+56=92 slices

8 0

2 years ago

Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class p

Dima020 [189]

Answer:

a) p-hat (sampling distribution of sample proportions)

b) Symmetric

c) σ=0.058

d) Standard error

e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

Step-by-step explanation:

a) This distribution is called the sampling distribution of sample proportions (p-hat).

b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.

This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.

c) The variability of this distribution, represented by the standard error, is:

$\sigma=\sqrt{p(1-p)/n}=\sqrt{0.16*0.84/40}=0.058$

d) The formal name is Standard error.

e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:

$\frac{\sigma_{90}}{\sigma_{40}}=\frac{\sqrt{p(1-p)/n_{90}} }{\sqrt{p(1-p)/n_{40}}}}= \sqrt{\frac{1/n_{90}}{1/n_{40}}}=\sqrt{\frac{1/90}{1/40}}=\sqrt{0.444}= 0.667$

If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

0 0

2 years ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago