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I am Lyosha [343]
3 years ago
9

On a Saturday the ratio of the amount of water used by Household A to the amount of water used by Household B was 13:5.Household

A used 260 gallons of water for that day.find the total amount of water used by the two household on that Saturday
Mathematics
1 answer:
Reptile [31]3 years ago
7 0
Use a proportion.

Let A = water used in household A.
Let B = water used in household B.

A to B = 13 to 5

A = 260 gal

260 to B = 13 to 5

260/B = 13/5

13B = 5 * 260

B = 100

A + B = 260 + 100 = 360

The total is 360 gallons.
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8. You recorded the time in seconds it took for 8 participants to solve a puzzle. The times were: 15.2, 18.8, 19.3, 19.7, 20.2,
kirill115 [55]

Answer:

Xbar = 20.813

Median = 19.95

5 - number summary :

Minimum = 15.2

Q1 = 19.05; Q2 = 19.95 ; Q3 = 21.95 ; maximum = 29.4

Standard deviation, s = 4.068

Step-by-step explanation:

Given the data:

X : 15.2, 18.8, 19.3, 19.7, 20.2, 21.8, 22.1, 29.4

Ordered data:

15.2, 18.8, 19.3, 19.7, 20.2, 21.8, 22.1, 29.4

The mean (xbar) :

Xbar = Σx / n ; n = sample size = 8

Xbar = 166.8 / 8 = 20.8125

Xbar = 20.813

The median :

1/2 *(n+1)th term

1/2(9) = 4.5

(4 + 5)th term / 2

(19.7 + 20.2) / 2 = 19.95

The five number summary :

Minimum = 15.2

Lower quartile (Q1) = 1/4(9) = 2.25 ; (2nd + 3rd) term / 2

(18.8 + 19.3) / 2 = 19.05

Median (Q2) = 19.95

Upper quartile (Q3) :

3/4(9) = 6.75 = (6th + 7th) / 2

(21.8 + 22.1) / 2 = 21.95

Maximum = 29.4

The standard deviation, s = sqrt[(x - xbar)²/ (n-1)]

Using calculator ;

s = 4.068

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3 years ago
hallie is trying to win the grand prize on a game show. Should she try her luck by spinning a wheel with 6 equal sections labele
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3 years ago
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Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo
Shkiper50 [21]

Answer:

a) P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335

b) P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452

c) P(X \geq 8) = 1-P(X

d) P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that X \sim Poisson(\lambda=4)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda=4  , Var(X)=\lambda=2, Sd(X)=2

a. Compute both P(X≤4) and P(X<4).

P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183

P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733

P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465

P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646

P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692

b. Compute P(4≤X≤ 8).

P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595

P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298

P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452

c. Compute P(8≤ X).

P(X \geq 8) = 1-P(X

P(X \geq 8) = 1-P(X

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

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Answer:

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