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2 years ago

As part of statistics project, math class weighed all the children who were willing to participate as shown below.

Mathematics

1 answer:

umka2103 [35]2 years ago

5 0

Answer:

Step-by-step explanation:

The key of this leaf stem-leaf plot is:

4 | 5 = 45

Therefore, the data points that represents children with weight that is less than 50 pounds would be as follows:

First row => 12, 13, 15, 15 15, 16, 16, 19

Second row => 23, 25

Third row => 33, 35, 37, 38, 39

Fourth row => 44, 45, 45

Therefore, the number of children that weigh less than 50 pounds = 18

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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:

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

7 0

3 years ago

PLS help 85% of 2500m

pishuonlain [190]

Answer:

2125m or 2km or 125m

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

3 0

2 years ago

If cos(x) = Three-fourths and tan(x) < 0, what is cos(2x)?

makvit [3.9K]

Step-by-step explanation:

The value of sin(2x) is \sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

How to determine the value of sin(2x)

The cosine ratio is given as:

\cos(x) = -\frac 14cos(x)=−

Calculate sine(x) using the following identity equation

\sin^2(x) + \cos^2(x) = 1sin

(x)+cos

(x)=1

So we have:

\sin^2(x) + (1/4)^2 = 1sin

(x)+(1/4)

\sin^2(x) + 1/16= 1sin

(x)+1/16=1

Subtract 1/16 from both sides

\sin^2(x) = 15/16sin

(x)=15/16

Take the square root of both sides

\sin(x) = \pm \sqrt{15/16

Given that

tan(x) < 0

It means that:

sin(x) < 0

So, we have:

\sin(x) = -\sqrt{15/16

Simplify

\sin(x) = \sqrt{15}/4sin(x)=

sin(2x) is then calculated as:

\sin(2x) = 2\sin(x)\cos(x)sin(2x)=2sin(x)cos(x)

So, we have:

\sin(2x) = -2 * \frac{\sqrt{15}}{4} * \frac 14sin(2x)=−2∗

∗

This gives

\sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

6 0

2 years ago

Read 2 more answers

A stack of one hundred fifty cards is placed next to a ruler, and the height of the stack is measured to be 5/8 inches.

DaniilM [7]

Divide 5/8 by 150, to see how thick each is, that'll give you 150 even pieces that add up to 5/8

$\bf \cfrac{\frac{5}{8}}{150}\implies \cfrac{\frac{5}{8}}{\frac{150}{1}}\implies \cfrac{5}{8}\cdot \cfrac{1}{150}\implies \cfrac{1}{8\cdot 30}\implies \cfrac{1}{240}$

5 0

3 years ago

Renna pushes the elevator button, but the elevator does not move. The mass limit for the elevator is 450 kg, but Renna and her l

pychu [463]

Answer:

$620-37.4p\leq 450$

Step-by-step explanation:

Let p be the number of identical packages.

We have been given that each package has a mass of 37.4 kg, so weight of p packages will be 37.4p.

The total mass of Renna and her load of identical packages is 620 kg.

We have been given that the mass limit for the elevator is 450 kg. This means that Renna can remove p packages from 620 kg such that 620 minus weight of p packages will be less than or equal to 450 kg.

We can represent this information in an inequality as:

$620-37.4p\leq 450$

Therefore, the inequality $620-37.4p\leq 450$ can be used to determine the number of packages, p, Renna could remove from the elevator to meet the mass requirement.

3 0

3 years ago

Read 2 more answers