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3 years ago

A survey of

Mathematics

1 answer:

Zina [86]3 years ago

7 0

Answer:

0.0208<p<0.0592

Step-by-step explanation:

-Given the sample size is 400 and the desired proportion is 16.

-The confidence interval can be determined as follows:

$CI=\hat p\pm z\sqrt{\frac{\hat p(1-\hat p}{n}}\\\\\hat p=x/n=16/400=0.04$

#We the use this proportion to find the CI at 95%:

$CI=0.04\pm 1.96\times \sqrt{\frac{0.04(1-0.04)}{400}}\\\\=0.04\pm 0.0192\\\\=[0.0208,0.0592]$

Hence, the 95% confidence interval is 0.0208<p<0.0592

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14 The straight line, L, has the equation y = 5 - 2x

lozanna [386]

Answer:

Step-by-step explanation:

(a) the co-ordinates of the point where the line crosses the y-axis,

y intercept occurs when x = 0

y = 5 - 2(0)

y = 5

(0, 5)

(b) the gradient of the line,

the general line formula is y = mx + b, where m is the slope and b is the y intercept.

y = 5 - 2x

y = -2x + 5

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Parallel lines have the same slope, but different y intercepts.

The simplest parallel line formula is

y = -2x + 0

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Step-by-step explanation:

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Plz help with math u will get Brainliest

emmasim [6.3K]

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Volume = Length × Width × Height

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3 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

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