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Bond [772]
2 years ago
5

Dominic has $600 in a savings account at the beginning of the summer. He wants to have at least $100 in the account by the end o

f the summer. He withdraws $25 each week for food, clothes, and movie tickets. Write an inequality that represents Dominic's situation. How many weeks can Keith withdraw money from his account
Mathematics
2 answers:
Free_Kalibri [48]2 years ago
8 0
He can withdraw 20 weeks in his account until to reach $100. Explanation: 25 + 25 + 25 + 25 is a hundred so keep counting on your fingers until you reach $100.
Ostrovityanka [42]2 years ago
7 0
Keith works 14 weeks to make $25
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411 because 438-27=411

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What would be the first step to solve for n in the equation 3n-7=30?
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The first step would be to add 7 to both sides.
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3 years ago
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The owner of Darkest Tans Unlimited in a local mall is forecasting the demand for November the one new tanning booth based on th
lorasvet [3.4K]

Answer:

The forecast for November is 235 if August's forecast was 145.

Step-by-step explanation:

The formula for calculating forecast using exponential smoothing is:

F_{t} = F_{t-1}  + \alpha (A_{t-1} - F_{t-1} )

Where Ft = New month forecast

           Ft-1 = Previous month forecast

           At-1 = Previous month actual value

            α = smoothing constant

We are given F₈ = 145 (forecast for August), A₈ = 200 (Actual Value for August), α = 2, and we need to compute the forecast for November. So, We will first calculate the forecast for September then October and then November, step-by-step.

So, forecast for September is:

F₉ = F₈ + α (A₈ - F₈)

    = 145 + 2*(200-145)

    = 145 + 2*55

F₉ = 255

Then, forecast for October is:

F₁₀ = F₉ + α (A₉ - F₉)

     = 255 + 2*(220-255)

     = 255 + 2*(-35)

F₁₀ = 185

The forecast for November is:

F₁₁ = F₁₀ + α (A₁₀ - F₁₀)

    = 185 + 2*(210 - 185)

F₁₁ = 235

8 0
3 years ago
A recent study of 26 city residents showed that the time they had lived at their present address has a mean of 10.3 years with t
svetlana [45]

Answer:

The 95% confidence interval is  9.15<  \mu < 11.45

Step-by-step explanation:

From the question we are told that

     The sample size is  n =  26

      The mean is  \= x =  10.3

       The standard deviation is  s =  3

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>   E =  1.96 *  \frac{3}{\sqrt{26} }

=>   E = 1.1532

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>   10.3  -1.1532 <  \mu < 10.3  +  1.1532

=>    9.15<  \mu < 11.45

7 0
3 years ago
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