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3 years ago

A container holds 16 pints of lemonade. How much is this in gallons

Mathematics

1 answer:

Sonja [21]3 years ago

7 0

Since one gallon is equal to 8 pints, 16 pints would equal to 2 gallons.

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ILL GIVE BRAINLIEST!!!!!! Please explain!! Thank youuuuu!

Natali5045456 [20]

Answer:

Table D

Step-by-step explanation:

you have to solve .3x = 1.5. this equals 5....since 5 is doubled to 10 you have to double 1.5 to 3.0

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3 years ago

Read 2 more answers

You are curious about the average number of yards Matthew Stafford throws for each game for the Detroit Lions. You randomly sele

Damm [24]

Answer:

The margin of error for this estimate is of 14.79 yards per game.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.093

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample.

You randomly select 20 games and see that the average yards per game is 273.7 with a standard deviation of 31.64 yards.

This means that $n = 20, s = 31.61$

What is the margin of error for this estimate?

$M = T\frac{s}{\sqrt{n}}$

$M = 2.093\frac{31.61}{\sqrt{20}}$

$M = 14.79$

The margin of error for this estimate is of 14.79 yards per game.

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3 years ago

BIG IDEAS MATH help me out please hurry

arsen [322]

I will but the picture is blurry

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2 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

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2 years ago

Two plants, plant A and B, are being grown in a science experiment. Plant A is 6 cm tall and growing at a rate of 4 cm/day. Plan

Irina18 [472]

By the second day of growth

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3 years ago