I need help big time! Thanks for anyone who can answer this question!

given the area of a rectangle is 14,628 square millimeters and the width is 12 millimeters find the length

HACTEHA [7]

A= l times w. A=14628 and w=12 so length equals 1219 millimeters

8 0

3 years ago

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What is 14.9 divided by 0.02

Lelechka [254]

The answer is 745. Doesn't math make sense ?

4 0

4 years ago

Can you help me with some steps to approach this proof .

Elis [28]

To prove two sets are equal, you have to show they are both subsets of one another.

• X ∩ (⋃ ) = ⋃ {X ∩ S | S ∈ }

Let x ∈ X ∩ (⋃ ). Then x ∈ X and x ∈ ⋃ . The latter means that x ∈ S for an arbitrary set S ∈ . So x ∈ X and x ∈ S, meaning x ∈ X ∩ S. That is enough to say that x ∈ ⋃ {X ∩ S | S ∈ }. So X ∩ (⋃ ) ⊆ ⋃ {X ∩ S | S ∈ }.

For the other direction, the proof is essentially the reverse. Let x ∈ ⋃ {X ∩ S | S ∈ }. Then x ∈ X ∩ S for some S ∈ , so that x ∈ X and x ∈ S. Because x ∈ S and S ∈ , we have that x ∈ ⋃ , and so x ∈ X ∩ (⋃ ). So ⋃ {X ∩ S | S ∈ } ⊆ X ∩ (⋃ ).

QED

• X ∪ (⋂ ) = ⋂ {X ∪ S | S ∈ }

Let x ∈ X ∪ (⋂ ). Then x ∈ X or x ∈ ⋂ . If x ∈ X, we're done because that would guarantee x ∈ X ∪ S for any set S, and hence x would belong to the intersection. If x ∈ ⋂ , then x ∈ S for all S ∈ , so that x ∈ X ∪ S for all S, and hence x is in the intersection. Therefore X ∪ (⋂ ) ⊆ ⋂ {X ∪ S | S ∈ }.

For the opposite direction, let x ∈ ⋂ {X ∪ S | S ∈ }. Then x ∈ X ∪ S for all S ∈ . So x ∈ X or x ∈ S for all S. If x ∈ X, we're done. If x ∈ S for all S ∈ , then x ∈ ⋂ , and we're done. So ⋂ {X ∪ S | S ∈ } ⊆ X ∪ (⋂ ).

QED

8 0

3 years ago

Nick dreamed that he was kidnapped

Sliva [168]

From the book "I Got You Babe"

8 0

3 years ago

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A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that wi

gulaghasi [49]

Answer:

a) 0.3585

b) 0.6415

c) 0.07548

Step-by-step explanation:

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 20

x = Number of successes required

p = probability of window frame with minor defect = 0.05

q = probability of window frame with no defect = 1 - 0.05 = 0.95

a) Probability that none of the frames picked will have defects

x = 0

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

P(X = 0) = ²⁰C₀ 0.05⁰ 0.95²⁰⁻⁰ = 0.3585

b) Probability that At least one will need adjustment = P(X ≥ 1) = 1 - P(X < 1) = 1 - P(X = 0) = 1 - 0.3585 = 0.6415

c) Probability that More than two will need adjustment = P(X > 2)

P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

So, we evaluate each of that using the binomial distribution formula, slotting in the appropriate variables, and sum it all up.

P(X > 2) = 1 - 0.9245 = 0.07548

7 0

4 years ago