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ella [17]
3 years ago
6

Number 1 to 20 are placed in a bag without replacing the first number what is the probability that the first number drawn will b

e odd and second one will be even
Mathematics
1 answer:
SOVA2 [1]3 years ago
7 0
Are there answer choices? if not for some reason i am going with 25 percent because there are 20 numbers and 10 are even and 10 are odd so the probability to get 2 odds is about 25%
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Determine whether the pair of actions results in independent or dependent events. Serena picks one marble out of a bag. Then, af
harina [27]

Answer:

B) Independent; the 1st marble selection will not affect the 2nd marble selection.

Step-by-step explanation:

When finding the probability of events in mathematics, we have both independent and dependent events.

Independent events are events that occur when the results of selection of the first events does not affect the results or outcomes of the second events.

Dependent events are the opposite of Independent events. They are events that occur when the results or outcomes obtained from the second events is affected by the results or outcomes from the selection of the first events.

From the question, we can see that the first event is she picked one marble from the bag. The second event is she replaced the marble before picking another marble. By doing this, the total number of possible outcomes for the probabilities of both events remains the same and they are unaffected.

Therefore, we can say that the two events are Independent because the 1st marble selection will not affect the 2nd marble selection.

4 0
3 years ago
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BRAINLIESTT ASAP! PLEASE HELP ME :)<br> thanks!
AVprozaik [17]

ΔACD and ΔASR are the same so we can write a ratio:

x/4 = 10/5

x/4 = 2

x/4 * 4 = 2 * 4

x = 8

Since AS is equal to x, they are both the same.

Therefore, the answer is [ x = 8, AS = 8 ]

Best of Luck!

6 0
3 years ago
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PLEASE HELP! WILL GIVE BRAINLIEST!
VMariaS [17]

Answer:

Answer: the equivalent form of 6(2x+3x+10) are 12x+18x + 60 and 30x+60.

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8 0
3 years ago
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A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

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

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

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

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

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_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0
3 years ago
How can you tell Is something is a unit rate
ArbitrLikvidat [17]

Unit rate is a ratio between two different units with a denominator of one. When we divide a fraction's numerator by its denominator, the result is a value in decimal form. For example: 8/4 = 2 and 3/6 = 0.5. When we write numbers in decimal form, we can write them as a ratio with one as the denominator.

For example, we can write 2 as 2/1, and 0.5 as 0.5/1. However, since that approach can be a little clumsy, we usually drop the one. That said, it's important to remember the one is there, especially when working with unit rates.

For instance, 8 miles/4 hours = 2 miles/hour. Notice again that, while we did not include the 1, we did include the unit 'hour' Miles per hour is a familiar expression, as are unit rates such as:

interest/amount invested

revolutions/minute

salary/year

Conversationally, the word ''per'' indicates we are using a unit rate.

4 0
3 years ago
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