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inn [45]
3 years ago
11

two bags each contain blue marbles and black marbles. the first contains 4 blue marbles and 6 black marbles and the second bag c

ontains 6 blue marbles and 6 black marbles. a marble is randomly drawn from each bag. what is the probability that both marbles are blue?. a. 1/5. b. 1/12. c. 3/10. d. 1/10
Mathematics
1 answer:
amid [387]3 years ago
4 0
There are two bags: a bag with 4 blue marbles and 6 black marbles and another one containing 6 blue marbles and <span>6 black marbles. the probability of getting a blue marble in the first bag is 6/10 or 3/5 while that on the second bag is 6/12 or 1/2. Hence the total probability of getting two marbles is c. 3/10. </span>
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Bryan gave a cashier $15 to pay for a sandwich and chips. He received $3.65 in change. If the sandwich cost $9.99, how much did
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Answer:1.36 for the chips

Step-by-step explanation:

15-9.99=5.01

5.01-3.65=1.36

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3 years ago
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Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog
lorasvet [3.4K]

Answer:

a) Standard Error = 0.010

b) 95% Confidence Interval = (0.0924 , 0.1316)

Step-by-step explanation:

a) The formula for Standard Error = √Sample Proportion (1 - sample proportion)/n

Standard Error = √p (1 - p)/n

We are told in the question that:

In a random sample of 400 people, 112 agree and 288 disagree. Estimate the standard error using 1000 samples

p = x/n

n = 1000 because we were told to use it instead of 400

x = number for people that agree = 112

p = 112/1000

p = 0.112

Standard Error = √p (1 - p)/n

= √0.112 (1 - 0.112)/1000

= √0.112 × 0.888/1000

= √0.099456 /1000

= √0.000099456

= 0.0099727629

Approximately to 3 decimal places = 0.010

Therefore, the standard error is 0.010

b) The Question above also asked that we solve for the 95% Confidence Interval

The formula =

p ± z × Standard Error

p = 0.112

z score for 95% confidence interval = 1.96

Standard Error = 0.010

Confidence Interval =

0.112 ± 1.96 × 0.010

= 0.112 ± 0.0196

0.112 - 0.0196

= 0.0924

0.112 + 0.0196

0.1316

Therefore, the 95% confidence interval = (0.0924 , 0.1316)

8 0
3 years ago
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The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in t
andreyandreev [35.5K]

Answer:

The test statistic is c. 2.00

The p-value is a. 0.0456

At the 5% level, you b. reject the null hypothesis

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly different than 3 minutes.

This means that the mean and the alternate hypothesis are:

Null: H_{0} = 3

Alternate: H_{a} = 3

The test-statistic is given by:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

Sample of 100 customers.

This means that n = 100

3 tested at the null hypothesis

This means that \mu = 3

The average length of time it took the customers in the sample to check out was 3.1 minutes.

This means that X = 3.1

The population standard deviation is known at 0.5 minutes.

This means that \sigma = 0.5

Value of the test-statistic:

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z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}

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The test statistic is z = 2.

The p-value is

Mean different than 3, so the pvalue is 2 multiplied by 1 subtracted by the pvalue of Z when z = 2.

z = 2 has a pvalue of 0.9772

2*(1 - 0.9772) = 2*0.0228 = 0.0456

At the 5% level

0.0456 < 0.05, which means that the null hypothesis is rejected.

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