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n200080 [17]
3 years ago
12

2. The picture shows a feeding trough that is shaped like a right prism. (a) The top of the feeding trough is open so that anima

ls can access its contents. List the area of each face of the feeding trough. (b) What is the surface area of the trough? Answer:
Mathematics
1 answer:
777dan777 [17]3 years ago
4 0
A)We have been given →
Hieght of trough = 5m
length of trough = 6m
Breadth of trough = 4m
so , As the top is opened for feeding animals.
Area of base = length× Breadth 
= (6×4) m²= 24 m²
Area of Front and back face = Hieght × length 
= (5×6) m² = 30m²
Area of side faces = hieght × Breadth
= (5×4) m²= 20 m²____________________________________
B)Total Surface area of trough will be given by formula - 2(lb+bh+hl)
= 2{(6×4)+(4×5)+(5×6)}
= 2{24+20+30}
= 2×74
= 148 m²
As we have been told top is open , so we will deduct top from Total surface Area to obtain Surface area →
(148-24) m²
(Area of top will be same as base area , that is 24m² as we have solved in A)
= 124m²
Answer→ Surface area of trough is 124m²
Read more on Brainly.com - brainly.com/question/4094485#readmore
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Answer:

(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.

(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.

(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.

(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.

The significance level for the test is <em>α</em> = 0.05.

The decision rule is:

If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 11.

The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.

The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₀ > 1.91) = 0.043.

The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>

The sample size is, <em>n</em> = 17.

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The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>

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The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.

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The null hypothesis is not rejected at 5% level of significance.

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The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 28.

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The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

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E(X) = \sum_{i=1}^n X_i P(X_I)

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