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Anton [14]
3 years ago
7

What is 29 fifths as a mixed number

Mathematics
2 answers:
oee [108]3 years ago
8 0
29/5 as a mixed number is 5 and 4/5
Dominik [7]3 years ago
5 0
Divide the 29 and 5 

and write any remainder 

29~5= 5.8 

5 4/5
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According to a polling​ organization, 24% of adults in a large region consider themselves to be liberal. A survey asked 200 resp
Varvara68 [4.7K]

Answer:

z=\frac{0.375 -0.24}{\sqrt{\frac{0.24(1-0.24)}{200}}}=4.47  

Now we can calculate the p value based on the alternative hypothesis with this probability:

p_v =P(z>4.47)=0.00000391  

The p value is very low compared to the significance level of \alpha=0.05 then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24

Step-by-step explanation:

Information given

n=200 represent the random sample taken

X=75 represent the number of people Liberal

\hat p=\frac{75}{200}=0.375 estimated proportion of people liberal

p_o=0.24 is the value that we want to test

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true proportion of adults liberal is higher than 0.24:

Null hypothesis:p \leq 0.24  

Alternative hypothesis:p > 0.24  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.375 -0.24}{\sqrt{\frac{0.24(1-0.24)}{200}}}=4.47  

Now we can calculate the p value based on the alternative hypothesis with this probability:

p_v =P(z>4.47)=0.00000391  

The p value is very low compared to the significance level of \alpha=0.05 then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24

6 0
3 years ago
Three vertices of a rectangle have coordinates (3,4), (5,−4), and (−7,−7). Determine the fourth vertex of the rectangle and then
Ket [755]
The fourth vertex would be at point (-9,1).
The slope from (5,-4) to (3,4) is -8/2 or -4.
So I did that with (-7,-7) to the fourth vertex to get the answer.
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3 years ago
GETT IT RIGHT AND GET BRAINLIEST
oksian1 [2.3K]

Answer:

18

Step-by-step explanation:

58 - |-40| =

58 -40=

18

7 0
4 years ago
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Express 15 kobo as a decimal of #3.00
arsen [322]

Answer:

15 kobocoin =30.143471

5 0
2 years ago
Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the
dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

I = \int \int_E \int zdV where E is bounded by the cylinder y^2 + z^2 = 81 and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and y^2 + z^2 = 81

∴ z^2 = 81 - y^2

z = \sqrt{81 - y^2}

Thus, z lies between 0 to \sqrt{81 - y^2}

GIven curve x = 0 and y = 9x

x =\dfrac{y}{9}

As such,x lies between 0 to \dfrac{y}{9}

Given curve x = 0 , x =\dfrac{y}{9} and z = 0, y^2 + z^2 = 81

y = 0 and

y^2 = 81 \\ \\ y = \sqrt{81}  \\ \\  y = 9

∴ y lies between 0 and 9

Then I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix}    ^ {\sqrt {{81-y^2}}}_{0} \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{(\sqrt{81 -y^2})^2 }{2}-0  \end {bmatrix}     \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{{81 -y^2} }{2} \end {bmatrix}     \ dxdy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0}    \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix}     \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81 \  y -y^3} }{18} \end {bmatrix}     \ dy

I = \dfrac{1}{18} \int^9_{y=0}  \begin {bmatrix}  {81 \  y -y^3}  \end {bmatrix}     \ dy

I = \dfrac{1}{18}  \begin {bmatrix}  {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}}  \end {bmatrix}^9_0

I = \dfrac{1}{18}  \begin {bmatrix}  {40.5 \ (9^2) - \dfrac{9^4}{4}}  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  3280.5 - 1640.25  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  1640.25  \end {bmatrix}

I = 91.125

4 0
3 years ago
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