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olga nikolaevna [1]
2 years ago
15

Suppose the probability that a randomly selected​ man, aged 55​ - 59, will die of cancer during the course of the year is StartF

raction 300 Over 100 comma 000 EndFraction . How would you find the probability that a man in this age category does NOT die of cancer during the course of the​ year?
Mathematics
1 answer:
sveticcg [70]2 years ago
7 0

Answer:

The probability that a man in this age category does NOT die of cancer during the course of the​ year is 0.997.

Step-by-step explanation:

Suppose the probability of an event occurring is P_{i}.

The probability of the given event not taking place is known as the complement of that event.

The probability of the complement of the given event will be,

1 - P_{i}

In this case an events <em>X</em> is defined as a man, aged 55​ - 59, will die of cancer during the course of the year.

The probability of the random variable <em>X</em> is:

P (X) = \frac{300}{100000}=0.003

Then the event of a man in this age category not dying of cancer during the course of the​ year will be complement of event <em>X, </em>denoted by<em> X</em>'.

The probability of the complement of event <em>X</em> will be:

P(X')=1-P(X)

          =1-0.003\\=0.997

Thus, the probability that a man in this age category does NOT die of cancer during the course of the​ year is 0.997.

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Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,
Bas_tet [7]

Answer:

a) \bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

Median = 25

b) Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) Midrange = \frac{70+13}{3}=41.5

d) Q_1 = \frac{20+21}{2} =20.5

Q_3 =\frac{35+35}{2}=35

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

Median = 25

Part b

The mode is the most repeated value on the dataset on this case would be:

Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

Midrange = \frac{Max+Min}{2}

And if we replace we got:

Midrange = \frac{70+13}{3}=41.5

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

Q_1 = \frac{20+21}{2} =20.5

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

Q_3 =\frac{35+35}{2}=35

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

6 0
3 years ago
Dominic took a 20-question online quiz. He received 5 points for every correct answer. For every
Gennadij [26K]

Answer:

X+20=2 i think this is a little above my head

8 0
2 years ago
Math Questions :3 thank uu &lt;3
coldgirl [10]
14.       1.5, 10    <- Answer
15.       5,1          <- Answer

Proof 14 

Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5               
                y = 10

Proof 15. 

Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer:  {x = 5                           y = 1
8 0
2 years ago
Please answer quickly I only have 5 mins lol. The ratio of students polled in 6th grade who prefer lemonade to Iced tea is 8:4,
zhannawk [14.2K]

Answer:

Let total students who prefer lemonade is 2a and who prefer ice tea is a.

Students in 6th grade=39

2a+a=39

3a=39

a=39/3

a=13

students who prefer lemonate is 13

and those who prefer ice tea is 2a=26

4 0
2 years ago
Simplify<br> (4i)(–5i)(6i)<br><br> –120<br><br> –120i(incorrect)<br><br> 120i<br><br> 120
professor190 [17]

Answer:

120i

Step-by-step explanation:

;i × i = -1....so hence

4i × (-5i) = -20 i^2...which is -20(-1) = 20....

then you multiply with 6i

20(6i) = 120i

4 0
3 years ago
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