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marishachu [46]
3 years ago
8

Least common multiple of 10 and 15

Mathematics
2 answers:
Aneli [31]3 years ago
4 0
The least common multiple of 10 and 15 is 5
bija089 [108]3 years ago
3 0
Least Common Multiple (LCM).

the multiples of 10: 0; 10; 20; 30; 40; ...
the multiples of 15: 0; 15; 30; 45; 60; ...

Answer: LCM(10; 15) = 30.
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Your family goes to a Southern-style restaurant for dinner. There are 6 people in your family.
swat32

Answer:

We conclude that:

  • The number of people who ordered the chicken dinner = 1
  • The number of people who ordered the steak dinner = 5

Step-by-step explanation:

Given that

  • Some order the chicken dinner for $14
  • some order the steak dinner for $17.
  • Total bill = $99

  • Let 'n' be the number of people who ordered the chicken dinner.
  • Let '6-n' be the number of people who ordered the steak dinner.

Thus, the equation becomes

14n + 17(6-n) = 99

14n + 102 - 17n = 99

-3n = 99 - 102

-3n = -3

Divide both sides by -3

n = 1

and

6-n = 6-1 = 5

Therefore, we conclude that:

  • The number of people who ordered the chicken dinner = 1
  • The number of people who ordered the steak dinner = 5
7 0
2 years ago
In rhombus ABCD, the diagonals AC and BDintersect at E. If AE=5 and BE=12, what is thelength of AB?
hram777 [196]

Diagonals that bisect each other are perpendicular and form a right angle.

AEB is a right triangle.

We can apply the Pythagorean theorem:

c^2 = a^2+b^2

Where c is the hypotenuse (longest side) and a and b the other 2 legs.

Replacing:

AB^2 = AE^2+BE^2

AB^2 = 5^2+12^2

AB^2 = 25+144

AB^2 = 169

AB=√169 = 13

Lenght of AB= 13

8 0
11 months ago
A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the prob
UkoKoshka [18]

Answer:

The probability that 8 mice are​ required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are​ required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

P(X=r)=^nC_r p^rq^{n-r}

Where, p is the probability of success p=\frac{2}{7}

q is the probability of failure q=1-p, q=1-\frac{2}{7}=\frac{5}{7}

n is total number of trials n=8

r=3

Substitute the values,

P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}

P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}

P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}

P(X=3)=0.2428

Therefore, the probability that 8 mice are​ required is 0.2428.

5 0
3 years ago
Please help ASAP! I will mark Brainliest! Please answer CORRECTLY! No guessing!
mezya [45]

Answer:

C. Exponential decay

Step-by-step explanation:

4 0
3 years ago
Craig just finished reading 120 pages of his history assignment if the assignment is 125 Pages what percent has Craig read so fa
diamong [38]
96% is the answer 100%/x%=125/120
(100/x)*x=(125/120)*x       - we multiply both sides of the equation by x
100=1.04166666667*x       - we divide both sides of the equation by (1.04166666667) to get x
100/1.04166666667=x
96=x
x=96
7 0
3 years ago
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