4^3 ⋅ 4^4 ⋅ 4^8 divided by 4^5 ⋅ 4

In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen

Lesechka [4]

Step-by-step explanation:

Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.

Type ni xi si

Graded 42 0.486 0.187

No-fines 42 0.359 0.158

To find - a. Formulate the above in terms of a hypothesis testing problem.

b. Give the test statistic and its reference distribution (under the null hypothesis).

c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.

Proof -

a.)

Hypothesis testing problem :

H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

b)

Test statistic :

$Z = \frac{x_{1} - x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1}} + \frac{s_{2}^{2} }{n_{2}} } }$

$Z = \frac{0.486 - 0.359 }{\sqrt{\frac{0.03496 }{42} + \frac{0.02496 }{42} } }$

$Z = \frac{0.127 }{\sqrt{0.001468}}$

$Z = \frac{0.127 }{0.0377}$

⇒Z(cal) = 3.3687

Z(tab) = 1.96

As Z (cal) > Z(tab)

So, we reject H0 at 5% Level of significance

p-value = 0.99962

Hence

There is significant difference in mean conductivity at the two materials.

5 0

3 years ago

See attached file, please help

Luba_88 [7]

System of Equations:
{-2x+2
{-1/3x-3

I used y=mx+b to find the equations

4 0

3 years ago

Suppose ABCD is a rhombus with AB = 12 inches. The midpoints of its sides are joined to form a quadrilateral. What type of quadr

MrRissso [65]

Answer:

A rectangle

Step-by-step explanation:

The given parameters are;

The quadrilateral ABCD is a rhombus;

The length of the sides of the rhombus = 12 inches

The lengths of the sides of a rhombus are equal

The opposite interior angles of the rhombus are equal

The length of each midpoint from the vertex = 12 in./2 = 6 in.

Therefore, we have;

The line joining the midpoints form a quadrilateral with the length of the opposite sides equal

The sum of the interior angles of the rhombus = 180°

From the diagram created with Microsoft Visio, we have;

4·a + 4·b + 360 = 4 × 180 = 360 + 360

4·a + 4·b = 360

a + b = 90°

We have;

The interior angles of the quadrilateral formed = x

a + b + x = 180° Sum of angles on a straight line

∴ x = 180° - (a + b) = 180° - 90° = 90°

x = 90°

Therefore, the interior angles of the quadrilateral formed = 90°

The quadrilateral formed having equal opposite sides, and all interior angle of 90° each is a rectangle

6 0

3 years ago

Two groups of staff at a store are making flowers to decorate the entrance. If Group A were to do it alone, it would take them 1

Tcecarenko [31]

Answer:

$\dfrac{6750}{7}$

Step-by-step explanation:

Let x be the number of flowers. If Group A were to make x flowers alone, it would take them 10 hours to complete the task, then they make $\dfrac{x}{10}$ flowers in an hour. If Group B were to make x flowers alone, it would take them 15 hours to complete the task, then they make $\dfrac{x}{15}$ flowers in an hour.

The groups decided to work together. They can make $\dfrac{x}{10}+\dfrac{x}{15}=\dfrac{x}{6}$ together in an hour and it takes them $\dfrac{x}{\frac{x}{6}}=6$ hours. Group B took a break for 1 hour and 40 minutes ( $1\dfrac{2}{3}$ hour), then Group B works

$6-1\dfrac{2}{3}=4\dfrac{1}{3}$ hours.

Thus,

$\dfrac{x}{10}\cdot 6-\dfrac{x}{15}\cdot 4\dfrac{1}{3}=300,\\ \\\dfrac{3x}{5}-\dfrac{13x}{45}=300,\\ \\14x=300\cdot 45=13500,\\ \\x=\dfrac{13500}{14}=\dfrac{6750}{7}.$