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elena-14-01-66 [18.8K]
2 years ago
9

On a popular television game show, a contestant must choose one of the five envelopes. One envelope contains the grand prize, a

car. Find the probability of not choosing a car. Show your solution.
Urgent! Please answer immediately. The answers that are nonsense will be reported. ​
Mathematics
1 answer:
lidiya [134]2 years ago
8 0

Answer:

4/5

Step-by-step explanation:

First, we know that we have 5 envelopes. So the denominator of the probability is going to be x/5. We then realize that only one envelope has the car, so the probability of that is 1/5. If we subtract 1/5 from the probability of choosing an envelope (5/5), we get 5/5-1/5=4/5.

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Dan takes an airplane to visit his parents. He pays the airline a total of $450. The cost of the ticket is $315. The airline als
vredina [299]

Answer:

9

Step-by-step explanation:

Start with the total amount Dan paid and subtract the part of the payment that is for the ticket.

450-315=135

Assuming that all of that $135 was for luggage fees and each pound over the limit costs $15, then the question is, how many times did Dan pay $15 until he'd paid the full $135 luggage fee.

So divide $135 by $15.

135/15=9

So his luggage weighted 59 pounds, which is 9 pounds over the limit.

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2 years ago
Please can someone help me with INTERIM CHECKPOINT Math problems
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A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after
lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                         P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of persons who were still employed primarily as engineers  = \frac{111}{200} = 0.555

           n = sample of graduates = 200

           p = population proportion of engineers

<em>Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                 significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } , 0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } ]

 = [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

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Answer:

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Step-by-step explanation:

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