Help me pleaseeeeeeee

Which of the following is the solution to | x| -5< -13 ?

sergey [27]

|x| - 5 < -13
|x| < -13 + 5
|x| < -8

No solution.

6 0

3 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

So

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

3 years ago

Find the sum of the first 20 terms of an arithmetic sequence with an 18th term of 8.1 and a common difference of 0.25.

il63 [147K]

The sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5

Given,

18th term of an arithmetic sequence = 8.1

Common difference = d = 0.25.

<h3>What is an arithmetic sequence?</h3>

The sequence in which the difference between the consecutive term is constant.

The nth term is denoted by:

a_n = a + ( n - 1 ) d

The sum of an arithmetic sequence:

S_n = n/2 [ 2a + ( n - 1 ) d ]

Find the 18th term of the sequence.

18th term = 8.1

d = 0.25

8.1 = a + ( 18 - 1 ) 0.25

8.1 = a + 17 x 0.25

8.1 = a + 4.25

a = 8.1 - 4.25

a = 3.85

Find the sum of 20 terms.

S_20 = 20 / 2 [ 2 x 3.85 + ( 20 - 1 ) 0.25 ]

= 10 [ 7.7 + 19 x 0.25 ]

= 10 [ 7.7 + 4.75 ]

= 10 x 12.45

= 124.5

Thus the sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5

Learn more about arithmetic sequence here:

brainly.com/question/25749583

#SPJ1

8 0

1 year ago

PLZ HELP ME PLZZZ 15 PTS Mr. Rodriguez asked 60 randomly chosen students from each grade level at Willowbrook School about their

vredina [299]

12/60 students chose science fiction

Approximately x/150 students prefer sf

60x = 1800

x = 30

30/150 students are assumed to prefer sf

30/150 = x/100

150x = 3000

x = 20

20/100 students are likely to prefer sf

Mr. Rodriguez made a reasonable estimate for the approximate percentage of students that prefer science fiction, because if 12/60 is equivalent to 30/150 which refers to the number of students who can be assumed to prefer science Fiction out of the whole school. Considering we need to identify what 30/150 as a percentage is, we can reduce it down to 1/5 to make I easier, then divide 1 by 5 to get .2
.2 as a percentage is 20%, so his inference was indeed reasonable.

(♥ω♥*)Brainliest Please(♥ω♥*)

8 0

3 years ago

Read 2 more answers

Someone pls help me with this I will make you brain

never [62]

Answer:

it is A the first one welcome pls mark as brainliest hope this helps

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers