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slamgirl [31]
3 years ago
14

Drawing a jack from a deck of cards, not replacing it, and then drawing another jack or examples of dependent variables.

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
7 0
<h2>Answer:</h2>

The given statement is a TRUE statement.

<h2>Step-by-step explanation:</h2>

Two events are said to be dependent if the happening of one event affect the happening of the second event.

In the given situation the given events are dependent.

  • Since in a deck of 52 playing cards there are 4 jack.

Hence, the probability of drawing a jack in the first draw is: 4/52

  • Now this drawn card is not replaced.

This means now in the second draw we have to chose a card from 51 remaining cards and also the number of jack that would be left are: 3

Hence, the probability of drawing a jack in second draw= 3/51

This means that the first event affect the happening of second.

              Hence, the events are dependent.

Basile [38]3 years ago
3 0
The answer to this question is true
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Russell works at a football concession stand selling drinks for $2 and bags of chips for $3
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Answer:

its 5

Step-by-step explanation:

2+3 = 5

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2 years ago
At what position on the number line is the red dot located?
Lostsunrise [7]

Answer:

square root of 7

Step-by-step explanation:

2 times 2 is 4

3 times 3 is 9

7 is the only option between 4 and 9

(this was more the easy method lol but hope this helps)

5 0
3 years ago
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2 years ago
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Carla is shopping for Storage Bags. At the grocery store, Brand A Storage Bags are priced $4.59 for 40 count, or Brand B Storage
vichka [17]

Answer:

Brand A

Step-by-step explanation:

Comment

The question is  one of finding out what the unit cost is.

Brand A: $4.59 has a count of 40

Brand B: $3.99 has a count of 30

Brand A

unit = 4.59 / 40 = 0.115

<em>Rounded: 0.12</em>

Brand B

unit = 3.99 / 30 = 0.133

<em>rounded: 0.13</em>

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8 0
1 year ago
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Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i
OLEGan [10]
<h2>Answer with explanation:</h2>

Let \mu be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

H_0:\mu=18\\\\H_a:\mu\neq18

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size :  n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : \overline{x}=18.8

Standard deviation : \sigma=2.8

Test statistic for population mean :-

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02

The p-value= 2P(z>2.02)=0.0433834

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0
3 years ago
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