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2 years ago

7

What is the volume of the following triangular prism?

1 answer:

wel2 years ago

8 0

I believe 252 yd^3

9 × 5 = 45

45 × 4 which is more than 126 I believe so, the higher number than that that is the only option is 252 yard^3

I hope this helps!

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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