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yanalaym [24]
3 years ago
8

What is the amount of liquid a water bottle bottle holds??

Mathematics
1 answer:
harkovskaia [24]3 years ago
8 0
If you look on the water bottle paper (It depends) you will find that it holds one of the amounts from 16.9<span> to 20 oz.

Hoped I helped!</span>
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It would be 14 for the perimeter
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According to Scarborough Research, more than 85% of working adults commute by car. Of all U.S. cities, Washington, D.C., and New
tamaranim1 [39]

Answer:

The 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}

Here the population standard deviation (σ) is not provided. So the confidence interval would be computed using the <em>t</em>-distribution.

The (1 - <em>α</em>) % confidence interval for population mean (<em>μ</em>) using the <em>t</em>-distribution is:

CI=\bar x\pm t_{\alpha /2,(n-1)}\frac{s}{\sqrt{n}}

Given:

\bar x=27.97\\s=10.04\\n=25\\t_{\alpha /2, (n-1)}=t_{0.01/2, (25-1)}=t_{0.005, 24}=2.797

*Use the <em>t</em>-table for the critical value.

Compute the 99% confidence interval as follows:

CI=27.97\pm 2.797\times\frac{10.04}{\sqrt{25}}\\=27.97\pm5.616\\=(22.354, 33.586)\\\approx(22.35, 33.59)

Thus, the 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

8 0
3 years ago
The fox population in a certain region has a continuous growth rate of 7 percent per year. It is estimated that the population i
rjkz [21]

Answer:

P(t) = 14300e^0.07t

Step-by-step explanation:

Let :

Population as a function of years, t = P(t) ;

Growth rate, r = 7%

Estimated population on year 2000 = Initial population = 14300

The given scenario can be modeled using an exponential function as the change in population is based in a certain percentage increase per period.

P(t) = Initial population*e^rt

P(t) = 14300*e^(0.07t)

P(t) = 14300e^0.07t

Where, t = number of years after year 2000.

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Solution: \left(2x^2+4x-3\right)-\left(2x^2+4x-3\right)=0

Steps:

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a, =2x^2+4x-3-\left(2x^2+4x-3\right)

-\left(2x^2+4x-3\right):\quad -2x^2-4x+3

=2x^2+4x-3-2x^2-4x+3

\mathrm{Simplify}\:2x^2+4x-3-2x^2-4x+3:\quad 0

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