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xeze [42]
2 years ago
5

Sarah is making a scale drawing of a painting that is 48 in. wide by 120 in. high. Her paper is 12 in. wide and 24 in. tall. She

decides to use the scale 1 in. = 4 in. Is this a reasonable scale?
Please show all work thank you.
Mathematics
2 answers:
romanna [79]2 years ago
8 0

Answer:

The scale is not reasonable .

Step-by-step explanation:

We are given that Sarah is making a scale drawing of a painting that is 48 in. wide by 120 in. high.

Her paper is 12 in. wide and 24 in. tall.

Scale :  1 in. = 4 in.

Since 1 inch = 4 inch

So, 12 inch = 12\times 4 = 48 inches

24 inch = 24\times 4 =96 inches

The given scale is not reasonable because the scale is true for the width but false for the height

Since the actual height is 120 inches and we are getting 96 inches using the given scale .

Hence the scale is not reasonable .

noname [10]2 years ago
3 0
48 / 12 = 4
120 / 24 = 5

This is not a reasonable scale because ratios of tall and wide are not proportional.

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Greetings!

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2 years ago
A store finds that its sales revenue changes at a rate given by S'(t) = −30t2 + 420t dollars per day where t is the number of da
allochka39001 [22]

Answer:

Step-by-step explanation:

Give the rate of change of sales revenue of a store modeled by the equation S'(t)= -30t^{2} + 420t. The Total sales revenue function S(t) can be gotten by integrating the function given as shown;

\int\limits {S'(t)} \, dt = \int\limits ({-30t^{2}+420t }) \, dt \\S(t) = \frac{-30t^{3} }{3}+\frac{420t^{2} }{2}\\  S(t)= -10t^{3} +210t^{2} \\

a) The total sales for the first week after the campaign ends (t = 0 to t = 7) is expressed as shown;

Given\ S(t) = -10t^{3} + 210t^{2}

S(0) = -10(0)^{3} + 210(0)^{2}\\S(0) = 0\\S(7) = -10(7)^{3} + 210(7)^{2}\\S(7) = -3430+10,290\\S(7) = 6,860

Total sales = S(7) - S(0)

= 6,860 - 0

Total sales for the first week = $6,860

b) The total sales for the secondweek after the campaign ends (t = 7 to t = 14) is expressed as shown;

Total sales for the second week = S(14)-S(7)

Given S(7) = 6,860

To get S(14);

S(14) = -10(14)^{3} + 210(14)^{2}\\S(14) = -27,440+41,160\\S(14) = 13,720

The total sales for the second week after campaign ends = 13,720 - 6,860

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8 0
3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales
andrezito [222]

Answer:

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=44 represent the sample mean

s=1.9 represent the sample standard deviation

n=40 sample size  

\mu_o =42 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 42, the system of hypothesis would be:  

Null hypothesis:\mu \leq 42  

Alternative hypothesis:\mu > 42  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=40-1=39  

Since is a one side test the p value would be:  

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

3 0
3 years ago
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