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2 years ago

I’m not sure what to do. Help please

Mathematics

1 answer:

Mars2501 [29]2 years ago

5 0

Answer:

put the answers in order and then the middle number is the mean

Step-by-step explanation:

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Help want to pass this!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

ddd [48]

Answer:

3.14 yd²

Step-by-step explanation:

Area = πr² = π(1)² = 3.14 yd²

6 0

3 years ago

Read 2 more answers

Solve for g.

Y_Kistochka [10]

Answer:

g =6/5 g=-5/2

Step-by-step explanation:

(5g − 6)(2g + 5) = 0

Using the zero product property

(5g − 6) =0 (2g + 5) = 0

5g-6+6 =0+6 2g+5-5=0-5

5g =6 2g =-5

5g/5 = 6/5 2g/2 =-5/2

g =6/5 g=-5/2

7 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

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

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

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

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

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

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

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

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

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

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

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

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

LESSON 23-1 PRACTICE

monitta

Answer:

Step-by-step explanation:

2 tuijnbb

6 0

3 years ago

Read 2 more answers

How do I simply this, thanks!

stepan [7]

(-a^3b^2*-a^-2b^-3)^-2/2a^2b^-3 = a^4b^9/2a^8b^4 =b^5/2a^4 so your answer is b^5/2a^4

7 0

3 years ago