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beks73 [17]
2 years ago
7

The number of people that can be safely carried in a lift depends on their mass. A lift will safely

Mathematics
1 answer:
vitfil [10]2 years ago
8 0

Answer:

Step-by-step explanation:

If a lift will carry 8 people safely with an average mass of 75kg each we multiply to find the weight the lift can safely carry.

8 x 75kg = 600kg

If the average mass of the people is 100kg each, we divide to find the number of people it can carry. We know the lift can carry 600kg.

600kg/100kg = 6

So the lift can carry 6 people if their average mass is 100kg.

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A store ships cans by weight. A small box can hold 3 to 5 pounds. A medium box can hold 5 to 8 pounds. A large box can hold 8 to
Strike441 [17]

Answer:

Step-by-step explanation: Uhh I have no idea how I got this right because I guessed but

small: 3-5 pounds 0.25+1.2+2.7

Medium: 5-8 pounds 0.25+0.25+1.2+1.2+1.2+1.2

Large 8-10 pounds 0.25+1.2+1.2+1.2+2.7+2.7

Hope I helped this is my first time answering a question have a good day everyone

6 0
2 years ago
What is the answer please
Marina86 [1]
Hey there!

Use The Pythagorean theorem. It’s a^2+b^2=c^2. Plug in 11 and 60. The equation would be 11^2+60^2=c^2. Simplify to get 121+3600=c^2. Simplify again to get 3721=c^2. Sqaure root each side to get c=61. The answer is the last one, c=61.

I hope this helps!
8 0
2 years ago
Write the subtraction fact 10-3 two ways
lisabon 2012 [21]
10-3 or 10+(-3) that should help
5 0
2 years ago
Read 2 more answers
oint Q is plotted on the coordinate grid. Point P is at (40, −20). Point R is vertically above point Q. It is at the same distan
JulsSmile [24]

Answer:

The Awnser Is C

Step-by-step explanation:

Hope This Helps! Have A Great Day

8 0
2 years ago
Read 2 more answers
Let Xn be the random variable that equals the number of tails minus the number of heads when n fair coins are flipped. What is t
Firlakuza [10]

Answer:

the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)

Step-by-step explanation:

If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1

therefore

Xn =X1-X2 = X1- (n-X1) = 2X1-n

thus

E(Xn) =∑ (2*X1-n) p(X1) =  2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0

the variance will be

σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) -  n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²

since

σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4

and

σ²(X1) = E(X1²) - [E(X1)]²

n/4 = E(X1²) - (n/2)²

E(X1²) = n(n+1)/4

therefore

σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²

= n - 2n² = n(1-2n)

σ²(Xn) = n(1-2n)

4 0
2 years ago
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