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algol [13]
2 years ago
9

A ladder is 4 feet and 1 inch tall. How tall is it in inches

Mathematics
2 answers:
Mice21 [21]2 years ago
4 0

Answer:

\huge\boxed{\sf  49\ inches}

Step-by-step explanation:

Length of the ladder = 4 feet 1 inch

We know that,

<h3>1 feet = 12 inches</h3>

So,

4 feets = 12 × 4 inches

4 feets = 48 inches

So,

<h3><u>Length of the ladder:</u></h3>

= 48 inches + 1 inch

= 49 inches

\rule[225]{225}{2}

castortr0y [4]2 years ago
3 0
<h2>SOLVING</h2>

\Large\maltese\underline{\textsf{A. What is Asked}}

If a ladder is 4 feet and 1 inch tall, how tall is it in inches?

\Large\maltese\underline{\textsf{B. This problem has been solved!}}

\boxed{\begin{minipage}{7cm} \\ The length of this ladder is 4 feet and 1 inch.\\The measure of one foot is 12 inches. \end{minipage}}

<u>Thus</u>,

\fbox{The measure of 4 feet is 12*4=48 inches.}

<u>We add 1 inch more.</u>

\bf{48+1=49\;inches}

\cline{1-2}

\bf{Result:}

               \bf{The\;ladder\;is\;49\;inches\;long.}

\LARGE\boxed{\bf{aesthetic\not1\theta\ell}}

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The height (in inches) of adult men in the United States is believed to be Normally distributed with mean μ . The average height
Jlenok [28]

Answer: a. 69.72 ± 1.37

Step-by-step explanation:

We want to determine a 90% confidence interval for the mean height (in inches) of adult men in the United States.

Number of sample, n = 25

Mean, u = 69.72 inches

Standard deviation, s = 4.15

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

69.72 ± 1.645 × 4.15/√25

= 69.72 ± 1.645 × 0.83

= 69.72 ± 1.37

The lower end of the confidence interval is 69.72 - 1.37 = 68.35

The upper end of the confidence interval is 69.72 + 1.37 = 71.09

5 0
3 years ago
11/4 hours 19/8 hours 2.6 hours witch is closest to 2 hours?
ExtremeBDS [4]

Answer:

\frac{19}{8} hours is closest to 2 hours.

Step-by-step explanation:

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In order to decide that, we will convert each of the given numbers into decimals and then we will compare.

\frac{11}{4} =2.750\\\frac{19}{8}=2.375\\2.6=2.600

Upon comparing 2.750, 2.375 and 2.600 we know that 2.375 is closest to 2. Therefore, out of the given three numbers, \frac{19}{8} is closest to 2.

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Find two power series solutions of the given differential equation about the ordinary point x = 0. y'' + xy = 0
nalin [4]

Answer:

First we write y and its derivatives as power series:

y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

(x+2)y′′+xy′−y=0

(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for  x  are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit  (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

Now we must set each coefficient, including constant term  =0 :

4a2−a0=0⟹4a2=a0

2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0

We would usually let  a0  and  a1  be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since  a0=4a2 , I’ll choose  a1  and  a2  as the two arbitrary constants. We can still express all other constants in terms of  a1  and/or  a2 .

an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)

a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2

a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2

a5=−(4⋅3)a4+2a32(5⋅4)=15!a2

a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2

We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all  n≥0 , except for  n=1 , since  a1  is an arbitrary constant independent of  a0  (and therefore independent of  a2 ).

Plugging back into original power series for  y , we get:

y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯

y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯

y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)

Notice that the expression following constant  a2  is  =4+  a power series (starting at  n=2 ). However, if we had the appropriate  x -term, we would have a power series starting at  n=0 . Since the other independent solution is simply  y1=x,  then we can let  a1=c1−3c2,   a2=c2 , and we get:

y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

Learn more about constants here:

brainly.com/question/11443401

#SPJ4

6 0
1 year ago
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