All categories

3 years ago

Which fraction is nearest in value

Mathematics

2 answers:

olganol [36]3 years ago

7 0

Answer:

it will be B took the test

Step-by-step explanation:

Lubov Fominskaja [6]3 years ago

4 0

Answer:

7/22

Step-by-step explanation:

You might be interested in

If sec0+tan0=p,then find the value of cosec0.

Ugo [173]

Answer:

Step-by-step explanation:

p=sec0+tan0

=1/cos0 + sin0/cos0

=(1+sin0)/cos0

square both sides

(1+sin)^2 /cos^2 = p^2

(1+sin)^2/(1 - sin^2 ) = p^2

(1+sin)^2/((1-sin)(1+sin)) = p^2

(1+sin)/(1-sin)=p^2

1+sin=p^2-p^sin

sin+p^2sin=p^2-1

sin(1+p^2)=(p^2-1)

sin=(1-p^2)/(1+p^2)

cosec=1/sin

=(1+p^2)/(1-p^2)

6 0

3 years ago

Given h(x) = 4x – 5, solve for a when h(x) = 7.

Elenna [48]

h(x)=4x-5

$\\ \sf\longmapsto h(7)$

$\\ \sf\longmapsto 4(7)-5$

$\\ \sf\longmapsto 28-5$

$\\ \sf\longmapsto 23$

3 0

3 years ago

Read 2 more answers

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

During the first six months of operation, Abdoul's lemonade stand made a profit of $50 per month. In the next three months it ha

MrMuchimi

profit, p = 6 × $50 = $300

losses, l = 120 + 30 + 200 = $350

overall = p - l = -$50

loss of $50

4 0

3 years ago

Number between 49 and 95 that's a multiple of 5,6 and 10

fiasKO [112]

5,10,15,20,25,30,35,40,45,50,55,60
6,12,18,24,30,36,42,48,54,60
10,20,30,40,50,60
The answer is 60

8 0

3 years ago