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Firlakuza [10]
3 years ago
14

A standard interior staircase has steps each with a rise (height) of 21 cm and a run (horizontal depth) of 24 cm. Research sugge

sts that the stairs would be safer for descent if the run were, instead, 29 cm. For a particular staircase of total height 5.04 m, how much farther would the staircase extend into the room at the foot of the stairs if this change in run were made?
Mathematics
1 answer:
Alisiya [41]3 years ago
6 0

Answer:

The staircase would extends 1.20 m if the changes were made.

Step-by-step explanation:

Acording tho the statement the total heigth is 5.04 m, that is 504 cm, each steps rises 21 cm. with this information we can calculte how many steps are in the whole staircase:

Number steps= \frac{Total heigth}{each step} \\Number steps= \frac{504}{21}  = 24 steps

To calculate how much the staircase will extend, whe have to multiply the horizontal depth times the number of steps.

using the run of 24 cm that gives us a lengt of 576 cm and if the changes were made the new length will be 696 cm.

So the staircase would extend an amount of (696 - 576), that is 120 cm, or 1.20 m, if the changes were made.

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pychu [463]
Mean: np=100\times0.20=20

Standard deviation: \sqrt{np(1-p)}=\sqrt{100\times0.20\times0.80}=4
4 0
3 years ago
What is 98.044 rounded to the nearest hundred
valentina_108 [34]
98.044 rounded to the nearest hundredth is:

98.04 because 4 is less than 5


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4 0
3 years ago
A square field has an area of 4823m²(show your work! ) ​
8_murik_8 [283]

Answer:

69.45 m to the nearest hundredth.

Step-by-step explanation:

I guees you want the length of one side.

A square has 4 sides of equal length and the area = s^2  where s = length of each side.

So the length of a side of a square of area 4823

= √(4823)

= 69.4478 m.

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2 years ago
Power Series Differential equation
KatRina [158]
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for y

\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0

which indeed gives the recurrence you found,

a_{n+3}=-\dfrac{n-3}{n+3}a_n

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that a_2=0, and substituting this into the recurrence, you find that a_2=a_5=a_8=\cdots=a_{3k-1}=0 for all k\ge1.

Next, the linear term tells you that 6a_0+6a_3=0, or a_3=a_0.

Now, if a_0 is the first term in the sequence, then by the recurrence you have

a_3=a_0
a_6=-\dfrac{3-3}{3+3}a_3=0
a_9=-\dfrac{6-3}{6+3}a_6=0

and so on, such that a_{3k}=0 for all k\ge2.

Finally, the quadratic term gives 6a_1-12a_4=0, or a_4=\dfrac12a_1. Then by the recurrence,

a_4=\dfrac12a_1
a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1
a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1
a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1

and so on, such that

a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}

for all k\ge2.

Now, the solution was proposed to be

y=\displaystyle\sum_{n\ge0}a_nx^n

so the general solution would be

y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots
y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)
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4 0
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LenKa [72]

Step-by-step explanation:

I am not sure, if something else is missing.

but given that one chart we can say that Craig had 2 fastest throws.

they were in the category 70-75 mph.

but we cannot say precisely which one was faster, and how fast it went.

it is also not clear what is the categorization of a border element (e.g. with an exact speed of 60 mph - is it in the 55-60 or in the 60-65 category ? or both ?).

I assume the upper limit of each interval is included, and the lower limit is excluded.

under this assumption we can say the fastest pitch was faster than 70 mph but slower than or equal to 75 mph.

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