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Gwar [14]
2 years ago
8

A computer shop charges 20 pesos per hour for first two hours and an additional 10 pesos per hour for each succeeding hour. Repr

esent your computer rental fee using the function R(t) where t is number of hours you spent on the computer
Mathematics
2 answers:
slamgirl [31]2 years ago
7 0

Answer:

R(t)=\left\{\begin{matrix}20t; & 0\leq t\leq 2 \\ 20+10t; & 2 < t\end{matrix}\right.

Step-by-step explanation:

Here, t represents the number of hours spent on using the computer,

Given,

The charge is 20 peso for first  two hours,

Thus, if R(t) represents the total fee for using the computer,

R(t)=20t for 0 ≤ t ≤ 2,

Also, after 2 hours the additional charges is 10 peso per hour,

For 2 hours the charges = 2 × 20 = 40 peso,

So, the charges for t hours for which t > 2,

R(t) = 40 + 10(t-2)

=40+10t-20

=20+10t

Hence, the required function that represents the computer rental fee is,

R(t)=\left\{\begin{matrix}20t; & 0\leq t\leq 2 \\ 20+10t; & 2 < t\end{matrix}\right.

SCORPION-xisa [38]2 years ago
4 0
R(t)=10t+20. This shows the first 20 pesos, and the additional 10 pesos for every hour. Hope this helps.
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The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution
oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!

Then, we can calculate P(X ≤ 5) as:

P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080

4 0
2 years ago
Zachary’s final project for college course took a semester to write and had 95,234 words.zachary wrote 35,295 words the first mo
kap26 [50]

Answer:

40,999

Step-by-step explanation:

95,234-35,295= 59,939 and then 59,939-19,240= 40,999

5 0
2 years ago
25x⁴+4x⁴y²+4y⁴ answer this​
vova2212 [387]

Answer:

(5x²+2y² + 2xy)(5x²+2y²-2xy)

Step-by-step explanation:

25x⁴+4x²y²+4y⁴= (5x²)²+2*5x²*2y²+(2y²)² - 16x²y²= (5x²+2y²)² - (4xy)²= (5x²+2y² + 2xy)(5x²+2y² - 2xy)

6 0
2 years ago
A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl
Citrus2011 [14]

Answer:

a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b)   95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

a)

<u>98% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-}+ t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

(664.14- 2.650\frac{297.29}{\sqrt{14} } , 664.14+ 2.650\frac{297.29}{\sqrt{14} } )

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

<u>95% of confidence intervals</u>

({s\sqrt{\frac{n-1}{X^{2} _{(\frac{\alpha }{2} ,n-1) } } } ,s\sqrt{\frac{n-1}{X^2_{\frac{1-\alpha }{2},n-1 } } }  )

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36     (check table)

X^2_{0.95,13} = 5.892    (check table)

(297.29. (\sqrt{\frac{14-1}{X^2_{0.05,13}  } } ),297.29(\sqrt{\frac{14-1}{X^2_{0.95,13} } } )

(297.29. (\sqrt{\frac{14-1}{22.36  } } ),297.29(\sqrt{\frac{14-1}{5.892 } } )

(214.91 , 441.53)

95% confidence interval for the standard deviation.

(214.91 , 441.53)

7 0
2 years ago
2 Rita bought three and forty-eight hundredths pounds of bananas at the store. How is this number written in expanded notation?
boyakko [2]

The cost of bananas = $ 3.48

This can be written as :                                                                                            

this number written in expanded notation as : (3 × 1) + (4 × 0.1) + (8 × 0.01)

3+0.4+0.08 = 3.48

Hence, 1st option is correct.                                                  

5 0
2 years ago
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