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Mnenie [13.5K]
2 years ago
6

The function f describes the value of a sculpture after t years. The function g describes the value of a painting by the same ar

tist after t years.
f(t) = 500(1.2)^t
g(t) = 380(1.15)^t

Find function h, such that h(t) = f(t) - g(t).

h(x) = 120(1.05)^t
h(t) = 880(1.35)^t
h(t) = 20[25(1.2)^t - 19(1.15)^t]
h(t) = 20[19(1.2)^t - 25(1.15^t]
Mathematics
2 answers:
leva [86]2 years ago
7 0
H(t)=ft)-g(t)
h(t)=500(1.2)^t-380(1.15)^t
Getting out 20 common factor:
h(t)=20[500(1.2)^t/20-380(1.15)^t/20]
h(t)=20[25(1.2)^t-19(1.15)^t]
Answer: Third option: h(t)=20[25(1.2)^t-19(1.15)^t] 
Anna11 [10]2 years ago
6 0
H(t) = f(t) - g(t)
= 500(1.2)^t - 380(1.15)^t
Taking out the greatest common factor, which is 20:
= 20[(25)(1.2)^t - (19)(1.15)^t]
This is the third choice.

Note that you cannot subtract the bases of the exponents, for example (1.2^t - 1.15^t) cannot be simplified into something like 0.05^t.
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Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in
Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                         P.Q.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of Americans who decide to not go to college = 48%

           n = sample of American adults = 331

           p = population proportion of Americans who decide to not go to

                 college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                        of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.645) = 0.90

P( -1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < \hat p-p < 1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

P( \hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

<u>90% confidence interval for p</u> = [ \hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } } , 0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } } ]

 = [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error =  Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }

              0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }

              \sqrt{n}  = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}

              \sqrt{n} = 54.79

               n = 54.79^{2}

               n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

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What are the coordinates of R' for the dilation D(0.5. p)(PQRS)?
RideAnS [48]

Answer:

Step-by-step explanation:

Three step process.

Step 1, convert Q, R, P, and S to coordinates measured from point P rather than from the origin by subtracting P from each point:

Q' = Q-P = (-3-(-3),2-(-3)) = (0,5)

R' = R-P = (1-(-3),2-(-3)) =(4,5)

P' = P-P = (-3-(-3),-3-(-3)) =(0,0)

S' = S-P = (1-(-3),-3-(-3)) = (4,0)

Step 2, Apply the dilation factor (0.5):

Q'' = (0.5)Q' = (0.5*0, 0.5*5) = (0,2.5)

R'' = (0.5)R' = _____

P''= (0.5)P' = _____

S''= (0.5)S' = _____

Step 3, Convert coordinates back to the origin (vice point P) by adding P to each point.  These are the dilated points.

Q''' = Q'' + P = (0+(-3),2.5+(-3)) = (-3, -0.5)

R''' = R'' + P = _____

P''' = P'' + P = _____

S''' = S'' + P = _____

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2 years ago
A swimmer can swim 50.2 meters in 1 minute. How far can he swim in half an hour? *
belka [17]

Answer:

1506meters

Step-by-step explanation:

1hour-60minutes

which means half an hour is 30minutes,so it's more like they are asking you to find how far the swimmer can swim in 30minutes

50.2-1min

x -30min

x=50.2×30

x=1506meters

6 0
2 years ago
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