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Alex787 [66]
2 years ago
14

A square has sides with a length of 5.8 inches. what is the effect on the perimeter of this square if the lengths of the sides a

re tripled?
Mathematics
1 answer:
Alla [95]2 years ago
4 0

By definition, the perimeter of a square is given by:

P = 4L

Where,

L: length of the sides of the square.

When the sides of the square are tripled, the equation changes as follows:

P '= 4 (3L)

Then, rewriting the equation we have:

P '= 3 (4L)

P '= 3P

Where,

P: perimeter of the original square.

Therefore, we can conclude that the perimeter is tripled.

Answer:

The perimeter of the square is tripled.

P '= 3P

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A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
Zielflug [23.3K]

Answer:

The 90% confidence interval using Student's t-distribution is (9.22, 11.61).

Step-by-step explanation:

Since we know the sample is not big enough to use a z-distribution, we use student's t-distribution instead.

The formula to calculate the confidence interval is given by:

\bar{x}\pm t_{n-1} \times s/\sqrt{n}

Where:

\bar{x} is the sample's mean,

t_{n-1} is t-score with n-1 degrees of freedom,

s is the standard error,

n is the sample's size.

This part of the equation is called margin of error:

s/\sqrt{n}

We know that:

n=28

\bar{x}=10.41

degrees of freedom = 27

1-\alpha = 0.90 \Rightarrow \alpha = 0.10

t_{n-1} = 1.703

s = 3.71

Replacing in the formula with the corresponding values we obtain the confidence interval:

\bar{x}\pm t_{n-1} \times s/\sqrt{n} = 10.41 \pm 1.70 \times 3.71/\sqrt{28} = (9.22, 11.61)

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2 years ago
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A company orders 6 boxed lunches from a deli for $41.40. If each boxed lunch costs the same amount, what is the unit cost of eac
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Answer:

248.40

Step-by-step explanation:

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According to a 2014 research study of national student engagement in the U.S., the average college student spends 17 hours per w
Nadusha1986 [10]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 17

For the alternative hypothesis,

µ < 17

This is a left tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 80,

Degrees of freedom, df = n - 1 = 80 - 1 = 79

t = (x - µ)/(s/√n)

Where

x = sample mean = 15.6

µ = population mean = 17

s = samples standard deviation = 4.5

t = (15.6 - 17)/(4.5/√80) = - 2.78

We would determine the p value using the t test calculator. It becomes

p = 0.0034

Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.

The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.

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2 years ago
In the equation "y=m+b," what does the "m" stand for​
Furkat [3]

The m in y=mx+b stands for the slope of the line

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The population P(t) of a species satisfies the logistic differential equation dP/dt= P(2-(P/5000)) where the initial population
Elis [28]
A logistic differential equation can be written as follows:
\frac{dP}{dt} = rP[1- \frac{P}{K}]

where r = growth parameter and K = carrying parameter.

In order to write you equation in this form, you have to regroup 2:
\frac{dP}{dt} = 2P[1- \frac{P}{10000}]

Therefore, in you case r = 2 and K = 10000

To solve the logistic differential equation you need to solve:

\int { \frac{1}{[P(1- \frac{P}{K})] } } \, dP =  \int {r} \, dt

The soution will be:

P(t) = \frac{P(0)K}{P(0)+(K-P(0)) e^{-rt} }

where P(0) is the initial population.

In your case, you'll have:

P(t) = <span>\frac{3E7}{3E3+7E3 e^{-2t} }

Now you have to calculate the limit of P(t).
We know that
</span>\lim_{t \to \infty}  e^{-2t} -\ \textgreater \  0  &#10;

hence,

\lim_{t \to \infty} P(t) =  \lim_{t \to \infty}  \frac{3E7}{3E3+0} =  10^{4}<span>

</span><span>

</span>
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