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3 years ago

Unit rate 1.29 for 12 rolls

Mathematics

1 answer:

murzikaleks [220]3 years ago

5 0

Answer:

.10705 cents (? your question doesn't specify if 1.29 is a dollar amount but I am assuming it is) per roll

Step-by-step explanation:

1.29/12=.10705

if 1.29 is cost in dollars round the number to $0.18

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At what point on the graph of y=2e^X -1 is the tangent line parallel to the line y= 5x -1

IgorLugansk [536]

Paralell has same slope
y=mx+b
m=slope
y=5x-1
when is the slope 5?

take the derivitive of 2e^x
f'(2e^x)=2f'(e^x)=2e^x

that is the slope
5=2e^x
divide both sides by 2
2.5=e^x
take the ln of both sides
ln2.5=x

when x=ln2.5 they are paralell

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3 years ago

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3 years ago

Marcel received an award for community service. The award included a check for $2,265. Three businesses each contributed the sam

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3 years ago

Read 2 more answers

Add or subtract the following polynomials

Rzqust [24]

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Step-by-step explanation:

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3 years ago

Consider the following data on x = rainfall volume (m^3) and y = runoff volume (m^3) for a particular location.

nydimaria [60]

Answer:

Step-by-step explanation:

From the given information:

x y xy x² y²

4 4 16 16 16

12 10 120 144 100

14 13 182 196 169

20 15 300 400 225

23 15 345 529 225

30 25 750 900 625

40 27 1080 1600 729

48 44 2112 2304 1936

55 38 2090 3025 1444

67 46 3082 4489 2116

72 53 3816 5184 2809

85 71 6035 7225 5041

96 82 7872 9216 6724

112 99 11088 12544 9801

127 101 12827 16129 10201

$\sum _{xi} = 805$ $\sum_{yi} = 643$ $\sum_{x_iy_I}= 51715$ $\sum_{x_i^2}= 63901$ $\sum_{y_i^2} = 42161$

The least-square regression equation is: $\hat y = b_o+b_1 x$

$b_1 = \dfrac{n \sum xy - ( \sum _x) ( \sum_y)}{n \sum x^2 - ( \sum x)^2}$

$b_1 = \dfrac{15(51715) - (803) (643)}{15(63901)-(805)^2}$

$b_1 = \dfrac{775725-516329}{958515-648025}$

$b_1 = \dfrac{259396}{310490}$

b₁ = 0.835440

∴ Slope term, b₁ = 0.835

$SS_{XX} = \sum x^2 - \dfrac{(\sum x)^2}{n}= 63901 - \dfrac{(805)^2}{15}=20699.33$

$SS_{yy} = \sum y^2 - \dfrac{(\sum y)^2}{n }= 42161 - \dfrac{(643)^2}{15}= 14597.73$

$SS_{xy} = \sum xy - \dfrac{(\sum x) (\sum y)}{n}= 51715 - \dfrac{(803)(643)}{15}= 17293.067$

$SST = SS_{yy}= 14597.73$

$SSR = \dfrac{SS_{xy}^2}{SS_{xx}}= \dfrac{17293.067^2}{20699.33}=14447.34$

SSE = SST - SSR = 14597.73 - 14447.34 = 150.39

The hypothesis test for the significance of $\beta_1$ is:

$H_o: \beta_1 = 0 \\ \\ H_1: \beta_1 \ne 0$

Significance level ∝ = 1 - 0.95 = 0.05

The sample slope $b_1$ = 0.835440

$Test \ statistic = t_{observed} = \dfrac{b_1-0}{\sqrt{\dfrac{SSE}{(n-2)SS_{xx}}}}$

$t_{o} = \dfrac{0.835440-0}{\sqrt{\dfrac{150.39}{(15-2)20699.33}}}$

$t_{o} = \dfrac{0.835440}{\sqrt{\dfrac{150.39}{269091.29}}}$

$t_o = 35.339$

Degree of freedom df = n - 2

df = 15 -2

df = 13

Using the Excel formula to determine the P_value.

$P-value = P(t, \Big|35.339 \Big|)$

P-value = 2 × t.dist(35.339,13,1)

P-value = 0.0000

P-value = 0

Critical value: $t_{critical} = t_{\alpha/2,df} = t_{0.05/2,13}= 2.160$

Rejection region: To reject $H_o$ ; if $\Big | t_o \Big | > t_c$

Decision: Since $\Big | t_o \Big | > t_c$ ; we reject $H_o$

Conclusion: There is enough evidence to conclude that the linear relationship between x & y

Thus; we reject $H_o$ & there is a useful linear relationship between x & y.

The 95% C.I for slope is given by the equation:

$=b_1 \pm t_{(\alpha/2,n-2)} \sqrt{\dfrac{SSE}{n-2}}\sqrt{\dfrac{1}{SS_{xx}} }$

$=0.835440 \pm 2.160 \sqrt{\dfrac{150.39}{15-2}}\sqrt{\dfrac{1}{20699.33} }$

= 0.835440 ± 2.160 (3.40124)(0.006951)

= 0.835440 ± 0.0511

= (0.835440 - 0.0511, 0.835440 + 0.0511)

= (0.78434, 0.88654)

= (0.784, 0.887) to three decimal places.

∴ 95% C.I of slope = $\mathbf{( 0.784 < \beta_1 < 0.887) \ to \ 3 \ d.p}$

6 0

3 years ago