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3 years ago

7

The number y of cups of flour used to make x loaves of bread is represented by the equation y=3x. Graph the equation. what are t

he points?

1 answer:

FromTheMoon [43]3 years ago

5 0

(-1,-3) , (-0.5,-1.5) , (0,0) , (0.5, 1.5) , (1,3)

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Two numbers totals -14, and their difference is 6. Find the two numbers

natita [175]

Answer:

The two numbers are -4 and -10.

Step-by-step explanation:

x+y=-14

x-y=6

----------

2x=-8

x=-8/2

x=-4

-4+y=-14

y=-14-(-4)=-14+4=-10

x=-4, y=-10.

4 0

3 years ago

How many people will 5 pitchers serve if pitcher serves 1/8 one person?

Marianna [84]

So if a person has 1/8 of a pitcher then you have 8 people per pitcher. Then if you have 5 pitchers you do 5 x 8 and get 40 so 40 people will be served. Hope this helps :)

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4 years ago

Please, I need help with this!! :(

hram777 [196]

PQS = 40 because if you look at the image above it half equals 20 and 20+20=40

4 0

3 years ago

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

<em>Part a: Sketch the deformed shape for α=0.03, β=-0.01 .</em>

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point A'(0+<em>(0.03)</em><em>(0),0+</em><em>(-0.01)</em><em>(0))</em>

Point A'(0<em>,0)</em>

Point B(x=1,y=0)

Point B'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point B'(1+<em>(0.03)</em><em>(1),0+</em><em>(-0.01)</em><em>(0))</em>

Point <em>B</em>'(1.03<em>,0)</em>

Point C(x=1,y=1)

Point C'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point C'(1+<em>(0.03)</em><em>(1),1+</em><em>(-0.01)</em><em>(1))</em>

Point <em>C</em>'(1.03<em>,0.99)</em>

Point D(x=0,y=1)

Point D'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point D'(0+<em>(0.03)</em><em>(0),1+</em><em>(-0.01)</em><em>(1))</em>

Point <em>D</em>'(0<em>,0.99)</em>

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

<em>Part b: Calculate the six strain components.</em>

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: <em>Find the volume change</em>

<em></em> $\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$ <em></em>

<em>Also the change in volume is 0.0197</em>

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

3 years ago

A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determinin

riadik2000 [5.3K]

Answer:

Option b - not significantly greater than 75%.

Step-by-step explanation:

A random sample of 100 people was taken i.e. n=100

Eighty of the people in the sample favored Candidate i.e. x=80

We have used single sample proportion test,

$p=\frac{x}{n}$

$p=\frac{80}{100}$

$p=0.8$

Now we define hypothesis,

Null hypothesis $H_0$ : candidate A is significantly greater than 75%.

Alternative hypothesis $H_1$ : candidate A is not significantly greater than 75%.

Level of significance $\alpha=0.05$

Applying test statistic Z -proportion,

$Z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Where, $\widehat{p}=80\%=0.80$ and $p=75%=0.75$

Substitute the values,

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}$

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.1875}{100}}}$

$Z=\frac{0.05}{0.0433}$

$Z=1.1547$

The p-value is

$P(Z>1.1547)=1-P(Z$

$P(Z>1.1547)=1-0.8789$

$P(Z>1.1547)=0.1241$

Now, the p-value is greater than the 0.05.

So we fail to reject the null hypothesis and conclude that the A is not significantly greater than 75%.

Therefore, Option b is correct.

7 0

3 years ago