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3 years ago

Can you please solve this

Mathematics

1 answer:

Vlad [161]3 years ago

5 0

Let the length = x and the width = y

For the length of fence we have x +x + y = 48 = 2x +y = 48

Rewrite as y = 48 - 2x

For the area we have x * y = 254

Replace y in the are equation:

x * 48 -2x = 254

Simplify the left side:

48x - 2x^2 = 254

Subtract 254 from both sides:

48x - 2x^2 - 254 = 0

Using the quadratic formula solve for x:

a = -2, b = 48 and c = -254

x = -48 + 4√17/-4

x = 16.123 m.

The length = x = 16.123 meters.

The width = 48 - (16.123*2) = 15.754 meters

Rounding to the nearest centimeter:

Length = 16.1 ( 16 meters and 1 cm)

Width = 15.8 ( 15 meters and 8 cm)

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Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa

Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation

n=580 represent the random sample taken

X represent the seafood sold in the country that is mislabeled or misidentified by the people

$\hat p=0.25$ estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

$\alpha=0.01$ represent the significance level (no given, but is assumed)

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204$

$0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296$

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

Part c

A government spokesperson claimed that the sample size was too small, relative to the billions of pieces of seafood sold each year, to generalize. Is this criticism valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

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4 years ago

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Step-by-step explanation:

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3 years ago

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liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7$

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

$\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55$

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

pretty much the same thing, we get the volume of the cone and its top, add them up.

$\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^$

8 0

3 years ago

Can you please solve this​

Can you please solve this