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Oksi-84 [34.3K]
3 years ago
12

1.11 People fit comfortably in a 5 feet by 5 feet area. Use this value to estimate the size of a crowd that is 5 feet deep on BO

TH sides of the street along a 5-mile section of a parade route.
(4 Points)
Mathematics
1 answer:
Oliga [24]3 years ago
3 0

Answer:

in 5ft by 5ft = 25ft^2, there can be 1 person

A normal street is 30ft wide

Then, if we want the width such that is 5 ft deep (from both sides)

Then we should subtract two times 5 ft:

30ft - 5ft - 5ft = 20ft.

Now the length is 5 miles

and 1 mile = 5280 ft

then 5 miles = 5*( 5280 ft) = 26,400 ft.

Then the area is:

20ft by 26,400 ft. =528,800  ft^2

And we know that in 25 ft^2 we can fit one person.

How many spaces of 25ft^2 we have in the 528,800 ft^2?

this is:

N =  528,800 ft^2/25ft^2 = 21,152

This means that in the street 21,152 people can fit comfortably.

The size of the crowd is  21,152 people.

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3 years ago
Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test t
bixtya [17]

Answer:

For this case we can find the critical value with the significance level \alpha=0.05 and if we find in the right tail of the z distribution we got:

z_{\alpha}= 1.64

The statistic is given by:

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

Replacing we got:  

z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86  

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of households with one person is significantly higher than 0.27

Step-by-step explanation:

We have the following dataset given:

X= 43 represent the households consisted of one person

n= 125 represent the sample size

\hat p= \frac{43}{125}= 0.344 estimated proportion of  households consisted of one person

We want to test the following hypothesis:

Null hypothesis: p \leq 0.27

Alternative hypothesis: p>0.27

And for this case we can find the critical value with the significance level \alpha=0.05 and if we find in the right tail of the z distribution we got:

z_{\alpha}= 1.64

The statistic is given by:

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

Replacing we got:  

z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86  

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of households with one person is significantly higher than 0.27

5 0
3 years ago
PLEASE HELP. I AM STRUGGLING.
Fittoniya [83]

(1.25x8)1.06=18.55

1.325x + 8.48=18.55

1.325x = 10.07

$7.60


4 0
3 years ago
to solve the system of equations below, Becca isolated x^2 in the firs equation and then substituted it into the second equation
bija089 [108]

The resulting equation if Becca isolated x² in the first equation and then substituted it into the second equation is (9-y²) / 25 - y²/36 = 1

<h3>Equation</h3>

x² + y² = 9

x²/25 - y²/36 = 1

From (1)

x² = 9 - y²

substitute x² = 9 - y² into (2)

x²/25 - y²/36 = 1

(9-y²) / 25 - y²/36 = 1

Therefore, the resulting equation is option C; (9-y²) / 25 - y²/36 = 1

Learn more about equation:

brainly.com/question/4172455

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4 0
2 years ago
Can you guys give me the steps for this plus the answer?
Alex_Xolod [135]

Answer:

x = -9

Step-by-step explanation:

First rearrange the terms.

-6x - 6x +3 = 111

Add like terms

-12x + 3 = 111

Subtract 3 from both sides

-12x = 108

Divide -12 on both sides

x = -9

3 0
2 years ago
Read 2 more answers
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