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azamat
2 years ago
5

Rahzel wants to determine how much gasoline he and his wife use in a month.He calculated that he used 78 1/3 gallons of gas last

month. Rahzel's wife used 41 3/8 gallons of gas last month. How much total gas did Rahzel and his wife use last month.Round your answer to the nearest hundredth. ( use a calculator to convert the fractions into decimals before calculating the sum or difference).
Mathematics
1 answer:
attashe74 [19]2 years ago
7 0
52. Rahzel wants to determine how much gasoline they had every month => He used 78 1/3 gallons of gas monthly => his wife used 41 3/8 gallons of gas last month How much is to total of gas that they both used. First let’s convert this fraction to decimal => 78 1/3 = 78.33 => 41 3/8 = 41.38 Now, let’s start adding. => 78.33 + 41.38 = 119.71 this is already rounded to the nearest hundredths.
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11 units is the correct answer to this question.

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2 years ago
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A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
2 years ago
Use Polya's four-step problem-solving strategy and the problem-solving procedures presented in this lesson to solve the followin
svp [43]

Answer:

a) 1+2+3+4+...+396+397+398+399=79800

b) 1+2+3+4+...+546+547+548+549=150975

c) 2+4+6+8+...+72+74+76+78=1560

Step-by-step explanation:

We know that a summation formula for the first n natural numbers:

1+2+3+...+(n-2)+(n-1)+n=\frac{n(n+1)}{2}

We use the formula, we get

a) 1+2+3+4+...+396+397+398+399=\frac{399·(399+1)}{2}=\frac{399· 400}{2}=399· 200=79800

b) 1+2+3+4+...+546+547+548+549=\frac{549·(549+1)}{2}=\frac{549· 550}{2}=549· 275=150975

c)2+4+6+8+...+72+74+76+78=S / ( :2)

1+2+3+4+...+36+37+38+39=S/2

\frac{39·(39+1)}{2}=S/2

\frac{39·40}{2}=S/2

39·40=S

1560=S

Therefore, we get

2+4+6+8+...+72+74+76+78=1560

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2 years ago
See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly se
REY [17]

Answer:

a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b) 0.0064% probability that all of the selected adults believe in reincarnation.

c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d) Since P(X \geq 5) < 0.05, 5 is a significantly high number of adults who believe in reincarnation in this sample.

Step-by-step explanation:

For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

n = 6, p = 0.2

a. What is the probability that exactly five of the selected adults believe in reincarnation?

This is P(X = 5).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015

There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b. What is the probability that all of the selected adults believe in reincarnation?

This is P(X = 6).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064

There is a 0.0064% probability that all of the selected adults believe in reincarnation.

c. What is the probability that at least five of the selected adults believe in reincarnation?

This is

P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564

There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?

5 is significantly high if P(X \geq 5) < 0.05

We have that

P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05

Since P(X \geq 5) < 0.05, 5 is a significantly high number of adults who believe in reincarnation in this sample.

5 0
2 years ago
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olga_2 [115]
I think 452cm^3
Voulme of one cylinder times 6
7 0
2 years ago
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