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2 years ago

Someone help me pls

Mathematics

1 answer:

aliya0001 [1]2 years ago

7 0

Step-by-step explanation:

that would be the right answer

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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

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

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

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

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

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

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

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

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

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

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

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

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

2 years ago

Please help me with this

MissTica

Answer:

For the question on the right, I assume b is the exponential factor, or growth factor.

Step-by-step explanation:

Linear:

$Y=aX + b\\(X, Y) = (0, 15) => b = 15\\(X,Y) = (1, 12) => a = -3\\Y = -3X + 15$

You can find the equation by using two points and the equation for a line.

Exponential:

$Y = a^{bX+c}\\(X, Y) = (0, 81) => a^{c} = 81 => a^{c} = 3^{4} => a=3, c=4\\(X, Y) = (1, 27) => a^{b+c} = 27 => a^{b}a^{c} = 27 => a^{b} = \frac{27}{81} = \frac{1}{3} => b=-1\\Y = 3^{-X+4}$

6 0

2 years ago

PLEASE HELP ASAP 5TH GRADE

natka813 [3]

Answer:

The answer would be B.

Step-by-step explanation:

Hope this helps.

5 0

2 years ago

Find lim x->2 for f(x)=

TiliK225 [7]

LHL in not equal to RHL , Therefore the limit does not exists , Option D is the answer.(none)

<h3>What is the limit of a function ?</h3>

The limit of a function at a certain point is the value that the function approaches as the argument of the function approaches the same point.

It is given that

lim x->2 for f(x)

$\lim_{x \rightarrow 2^-}f(x) = \lim_{x\rightarrow 2^+}f(x)$

f(x) = 2x+1 x ≤2

f(x)= x² , x >2

When both the function tends to 2

Left Hand Limit

f(x) = 2 *2 +1

f(x) = 5

Right Hand Limit

f(x) = x² ,

f(x) = 4

LHL in not equal to RHL , Therefore the limit does not exists.

To know more about Limit of a Function

brainly.com/question/7446469

#SPJ1

8 0

1 year ago

Which expression best estimates 11 1/5 divided by 2 3/4? <br><br> 4<br><br> 6<br><br> 11<br><br> 22

Maksim231197 [3]

11 1/5 = (11*5+1)/5=56/5
2 3/4= (2*4+3)/4=11/4
11 1/5 : 2 3/4=56/5:11/4=56/5*4/11=(56*4)/(5*11)=224/55≈4.07≈4
Answer is 4

5 0

2 years ago