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3 years ago

Find the critical value tc for the confidence level cequals 0.95 and sample size nequals 19.

1 answer:

8 0

The critical values are fixed values which can be obtained from the standard t crit tables. We can find the critical value tc when the degrees of freedom and confidence level are given. The degrees of freedom is simply:

Degrees of Freedom = Number of samples – 1

Degrees of Freedom = 19 – 1

Degrees of Freedom = 18

From the t critical tables, the tc corresponding to Degrees of Freedom equal to 18 and Confidence level equal to 95% is as follows:

tc = 2.101

Therefore this means that any t value beyond 2.101, we already reject the null hypothesis.

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Tickets to a baseball game cost $18.95, $12.95, or $9.95. A hot dog and soda combo costs $5.50. The Madison family is having a r

I am Lyosha [343]

Hmmm Interesting: 189.5+129.5+99.5+53
I got 471.5 Dollars....

Because Look If they buy 10 tickets each category it be 18.95 times 10= 189.5
then 12.95 times= 129.5 , Then 9.95 times= 99.5.... Also if they plan to get 30 combos it be also 5.50 times 30,,, You add the answers and it will give you 471.5 dollars. You can try doing it in a caculator (:

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3 years ago

Read 2 more answers

Plz help one question

Vladimir [108]

z° = 80° (Being corresponding angles)

y° + 80° = 180° (Being linear pair of angles)

=> y° = 180° - 80°

=> y° = 100°

x° = y° (Being interior opposite angles)

=> x° = 100°

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3 years ago

The rectangular prism below has a length of 18 inches, a width of 24 inches, and a height of 12 inches.

andrew11 [14]

Answer:

1872

Step-by-step explanation:on edge

3 0

4 years ago

in a recent semester at a local university 520 students enrolled in both general chemistry and calculus 1. Of these students 88

IceJOKER [234]

Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that

P(Ch) = 88/520

P(C) = 76/520

P(Ch and C) = 31/520

and we want to find P(Ch or C).

Using the inclusion/exclusion principle, we have

P(Ch or C) = P(Ch) + P(C) - P(Ch and C)

P(Ch or C) = 88/520 + 76/520 - 31/520

P(Ch or C) = 133/520

6 0

3 years ago

A banks loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviati

Alona [7]

Answer:

2.87%

Step-by-step explanation:

We have the following information:

mean (m) = 200

standard deviation (sd) = 50

sample size = n = 40

the probability that their mean is above 21.5 is determined as follows:

P (x> 21.5) = P [(x - m) / (sd / n ^ (1/2))> (21.5 - 200) / (50/40 ^ (1/2))]

P (x> 21.5) = P (z> -22.57)

this value is very strange, therefore I suggest that it is not 21.5 but 215, therefore it would be:

P (x> 215) = P [(x - m) / (sd / n ^ (1/2))> (215 - 200) / (50/40 ^ (1/2))]

P (x> 215) = P (z> 1.897)

P (x> 215) = 1 - P (z <1.897)

We look for this value in the attached table of z and we have to:

P (x> 215) = 1 - 0.9713 (attached table)

P (x> 215) =.0287

Therefore the probability is approximately 2.87%

4 0

4 years ago