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2 years ago

36 is 12% of what number?

Mathematics

2 answers:

svlad2 [7]2 years ago

8 0

X-> the number

12% • x = 36
12/100 • x = 36
3/25 • x = 36
x= 300

Alex777 [14]2 years ago

7 0

4.32 is the answer to ur question if u need to round then it would be 5

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Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug

gulaghasi [49]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems

In C(5,3)=10ways, this can be achieved

In part a:

Space sample chooses 3 of a 5 systems

$(M_1, \ M_2,\ M_3),$ $(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),$ $(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}$

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort

Outcomes probable =

$(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6$

$\to p(A)=\frac{6}{10}\\\\$

$=0.6$

In point c:

B = highest effort that is $M_5$

Potential result=

$(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6 \\\\$

$\to B= \frac{6}{10} \\\\$

$=0.6$

$\to P(B)=10$

In point d:

$\to \ A \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)$

$\to A (A \ intersection \ B) = \frac{3}{10} \\\\\ \ \ \ \ \$

$=0.3$

In point e:

$\to (A \cup B) = (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9$ $\to P(A \cap B)=\frac{9}{10}$

$= 0.9$

In point f:

$\to (A\cap B) = \frac{3}{10}$

$= 0.3$

In point g:

$\to (A \cup B) = \frac{7}{10}$

$=0.7$

In point h:

$\to p(A \cap B) = 0.3 \neq 0$

8 0

3 years ago

PLEASE HELP ASAP!!! <br> Thank you!

Marrrta [24]

The answer is 380 cm. I first found the area of the triangles (14x10x0.5), then I multiplied that by four (because there are 4 triangles total) getting 280. Then I added the area of the base (10x10=100) to 280. Good luck my bud

4 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

2 years ago

Theresa has two brothers,Paul and Steve, who are both the same height. Paul says he is 16 inches shorter than 1 1/2 times Theres

shepuryov [24]

Answer:

Theresa's height is 60 inches

Step-by-step explanation:

The given parameters are;

The height of Paul = The height of Steve

Paul's height = $1\dfrac{1}{2}$ times Theresa's height - 16 inches

Steve's height = $1\dfrac{1}{3}$ times Theresa's height - 6 inches

Let 'T' represents Theresa's height, we have;

Paul's height = $1\dfrac{1}{2}$ ×T - 16

Steve's height = $1\dfrac{1}{3}$ ×T - 6

Paul's height = Steve's height

Therefore;

$1\dfrac{1}{2}$ ×T - 16 = $1\dfrac{1}{3}$ ×T - 6

$1\dfrac{1}{2}$ ×T - $1\dfrac{1}{3}$ ×T = 10

$#\dfrac{3}{2}$ ×T - $\dfrac{4}{3}$ ×T = 10

$\dfrac{1}{6}$ ×T = 10

T = 10 × 6 = 60

Theresa's height, T = 60 inches

5 0

3 years ago

One eighth of a number is ten less than one third of the number. Find the number.

GarryVolchara [31]

X/8 - 10 = x/3

bring the values with the variable on one side

x/3 - x/8 = 10

find a common denominator and subtract these fractions

8x/24 - 3x/24 = -10

now subtract fractions

8x - 3x = 5x

5x/24 = -10

multiply both sides by 24

5x = -240

divide both sides by 5

x = -48

the number is -48

7 0

3 years ago

Read 2 more answers