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denis23 [38]
4 years ago
8

20 POINTS PLS HELP ASAP!!!

Mathematics
1 answer:
bogdanovich [222]4 years ago
6 0

Answer:

  • 10+x
  • 16-2x
  • -2x^2 -4x +160 ≥ 130
  • $12 or $13
  • $3

Step-by-step explanation:

If x represents the number of $1 increases from $10 in the cost of the buffet, then the cost per customer is ...

  c(x) = 10 +x

If n represents the number of customers for a given number of $1 increases (x), then we have ...

  n(x) = 16 -2x

The revenue will be the product of these:

  r(x) = c(x)n(x) = (10 +x)(16 -2x)

  r(x) = -2x^2 -4x +160

Then the desired inequality is ...

  r(x) ≥ 130

  -2x^2 -4x +160 ≥ 130

The solution to this is ...

  -2x^2 -4x +30 ≥ 0 . . . .subtract 130

  x^2 +2x -15 ≤ 0 . . . . . . divide by -2

  (x +5)(x -3) ≤ 0

The factors both have the same sign (hence a positive product) for x < -5 or x > 3. One of them is negative in the interval (-5, 3), so that is the solution to the inequality.

  -5 ≤ x ≤ 3

Noah could charge $12 or $13 and maintain his revenue.

The maximum possible increase that maintains Noah's revenue is $3. This is the upper end of the solution space for the inequality.

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(7.4*105)-(8.9*103) = (777)-(916.7) =(-139.7), not sure, did you write the problem wrong? If the total of the company is less than what the account dept used, something is off.  
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4 years ago
A random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph. From past information, it is kn
loris [4]

Answer:

95 percent confidence interval for μ is determined by

(60.688 ,  69.312)

a) be the same

Step-by-step explanation:

<u>Step (i)</u>:-

Given data a random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph.

The sample size is n= 121

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

<u>Step (ii) :</u>-

Given  The 95 percent confidence interval for μ is determined as

(61.08, 68.92)

we are to reduce the sample size to 100 (other factors remain unchanged) so

given The sample size is n= 100

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

<u>Step (iii)</u>:-

95 percent confidence interval for μ is determined by

(x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )

(65 - 1.96\frac{22}{\sqrt{100} } , 65 + 1.96 \frac{22}{\sqrt{100} } )

(65 - 4.312 ,65 +4.312)

(60.688 ,  69.312) ≅(61 ,69)

This is similar to (61.08, 68.92) ≅(61 ,69)

<u>Conclusion</u>:-

it become same as (61.08, 68.92)

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3 years ago
Read 2 more answers
The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth
Art [367]

Answer:

a) N_0=250\; k=0.15

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies t=0. And N(0)=N_0. The process is:

N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250

b) After 2 days means t=48. So, we just replace and operate:

N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}

c) N(t_1)=4000; \;t_1=?

N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\  \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h

d) t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)

N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h

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3 years ago
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