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2 years ago

Tell whether the ratios form a proportion. 3/4 and 24/18

Mathematics

1 answer:

Tresset [83]2 years ago

3 0

Answer:

Step-by-step explanation:

No. When you cross multiply the numbers should be equal.

So (3)(18) = (4)(24) but they are not equal so it is not a proportion

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A woman walks 5 rounds along the border of a square field of area 49². Find the total distance she walks in m.

Luba_88 [7]

You need to estimate the square root of 200 ft². Rewrite 200 as 2 x 100, then square root of (2*100) is:

10*sqrt(2). The sqrt(2) = 1.414, so the length of each side of the square will be 10*1.414 = 14.14 ft or about 14 ft.

4 0

2 years ago

Read 2 more answers

X + 2y - 3z = 11<br> 2x + y + z = -1<br> x - 3y - z=-8

Sav [38]

Answer:

x = -1 y = 3 and z =2

Step-by-step explanation:

that is the solution above

5 0

1 year ago

Help please and show the steps

marshall27 [118]

Answer:

-sin(2A)

Step-by-step explanation:

1-(sin(A)+cons(A))^2

1-(sin(A)^2+2sin(A)cos(A)+cos(A)^2) :use FOIL to get it

1-(1+sin(2A))

1-1-sin(2A)

-sin(2A)

8 0

3 years ago

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

balu736 [363]

Answer:

$P(X \ge 1) = 0.509$

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$