Estimate the quotient for the following problem 645 divided by 69

Simplify: 39w^5 - 13w^5

Alenkinab [10]

Answer:

hello :

Step-by-step explanation:

39w^5 - 13w^5 = 13w^5 (3 - 1) = 26w^5

5 0

3 years ago

What is each percent of change?

Evgen [1.6K]

Answer:

there is a 50% increase of change from 200 to 300. What do you mean by total cost or sale price?

Step-by-step explanation:

7 0

2 years ago

Nolini says that if the denominator is more than twice the numerator, the fraction can always be replaced with a 0. Is she corre

Len [333]

Answer:

yes

Step-by-step explanation:

It depends on how accurate you want your answer to be, but basically yes, imagine we set our numerator to be x, and our denominator to be 2x, if we built our fraction we will get that:

$\frac{x}{2x}=\frac{1}{2}$

If the denominator of a fraction is exactly twice the numerator, then the fraction will be simplified to 1/2 or 0.5.

Now, the greater the denominator is, the closer the fraction will get to zero.

Take for example I had an fraction like this:

$\frac{3}{6}=0.5$

which approximates to 0. If we made the denominator bigger, let's say 7, we would get:

$\frac{3}{7}=0.43$

notice the answer is closer to 0 this time, so it's valid to round it to zero. If we made the denominator greater, let's say 25, we would get:

$\frac{3}{25}=0.12$

and so on. So it is true that if the denominator of a fraction is greater than twice the numerator, we can always replace the fraction with a 0 (depending on how accurate you want your answer to be).

5 0

3 years ago

More help hahhahahha

Marat540 [252]

Answer:

s<u><</u>30,000

t<u>></u>50

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Trainees must complete a specific task in less than 2 minutes.Consider the probability density function below for the time ittak

grigory [225]

Answer:

1) P(X<1) = 0.5850

2) P(X>1) = 0.41500

3) P(0.68 < X < 1) = 0.0013

4) E(X) = 0.8867 minute

5) E(X²) = 1.1064

6) Var(X) = 0.3202

Step-by-step explanation:

Probability density function = f(x) = 0.67 - 0.17x (for 0 < x < 2)

1) The probability a trainee will complete the task inless than 1 minutes is given as P(X<1)

And,

P(X<1) = 1 - P(X≥1)

P(X≥1) = ∫²₁ f(x) dx = ∫²₁ (0.67 - 0.17x) dx = [0.67x - 0.085x²]²₁ = [0.67(2) - 0.085(2²)] - [0.67(1) - 0.085(1²)] = (1) - (0.585) = 0.415

P(X<1) = 1 - P(X≥1) = 1 - 0.415 = 0.5850

2) The probability that a trainee will complete the task inmore than 1 minutes is given as P(X>1)

And

P(X>1) = 1 - P(X≤1)

P(X≤1) = ∫¹₀ f(x) dx = ∫¹₀ (0.67 - 0.17x) dx = [0.67x - 0.085x²]¹₀ = (0.67(1) - 0.085(1²)) = 0.67 - 0.085 = 0.585

P(X>1) = 1 - P(X≤1) = 1 - 0.585 = 0.4150

3) The probability it will take a trainee between 0.68minutes and 1 minutes to complete the task. P(0.68 < X < 1)

P(0.68 < X < 1) = P(X<1) - P(X>0.68)

P(X<1) = 0.5850 (from question number 1)

P(X>0.68) = 1 - P(X≤0.68)

P(X≤0.68) = ∫⁰•⁶⁸₀ f(x) dx = ∫⁰•⁶⁸₀ (0.67 - 0.17x) dx = [0.67x - 0.085x²]⁰•⁶⁸₀ = (0.67(0.68) - 0.085(0.68²)) = 0.4556 - 0.039304 = 0.416296 = 0.4163

P(X>0.68) = 1 - P(X≤0.68) = 1 - 0.4163 = 0.5837

P(0.68 < X < 1) = P(X<1) - P(X>0.68) = 0.585 - 0.5837 = 0.001296 = 0.0013 to 4d.p

4) Expected value = E(X) = Σ xᵢpᵢ = ∫²₀ x f(x) dx (that is, a sum of all the products of possible values and their respective probabilities)

∫²₀ x f(x) = ∫²₀ x(0.67 - 0.17x) = ∫²₀ (0.67x - 0.17x²) dx = [0.335x² - 0.0567x³]²₀ = [0.335(2²) - 0.0567(2³)] = 1.34 - 0.4533 = 0.8867 minute.

E(X) = ∫²₀ x f(x) = 0.8867 minute

5) E(X²) = Σ xᵢ²pᵢ = ∫²₀ x² f(x)

∫²₀ x² f(x) = ∫²₀ x²(0.67 - 0.17x) = ∫²₀ (0.67x² - 0.17x³) dx = [0.2233x³ - 0.0425x⁴]²₀ = [0.2233(2)³ - 0.0425(2)⁴] = 1.7864 - 0.68 = 1.1064

6) Variance = Var(X) = Σx²p − μ²

where μ = E(X) = 0.8867 minute

Σx²p = ∫²₀ x² f(x) = 1.1064

Var(X) = 1.1064 - 0.8867² = 0.3202

5 0

2 years ago