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3 years ago

Dante's bath towel is 1 yard long. What measure is equivalent to 1 yard, in inches?

Mathematics

2 answers:

Leya [2.2K]3 years ago

7 0

Answer: 36 inches

Step-by-step explanation:

Because 1 yard has 3 feet in it and 1 feel = 12 inches so multiply it by 3

dybincka [34]3 years ago

4 0

Answer: 36 inches

Step-by-step explanation:

3 feet=36 inches

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Show all work to multiply (6+square root -64)(3-square root -16)

Hunter-Best [27]

Answer:

Step-by-step explanation:

- 8i = √-64

8i + 6

- -4i = -√-16

-4i + 3

[8i + 6][-4i + 3] → FOIL

>> -32i² + [24i - 24i] + 18

↑ ↑

-1 0

>>> 32 + 18 = 50

Information on Imaginary Numbers

√-1 = i

-1 = i²

-i = i³

1 = i⁴ [And every other exponent that is a multiple of 4; this cycle then repeats itself every time you go up one number at a time]

If you are ever in need of assistance, do not hesitate to let me know by subscribing to my You-Tube channel [USERNAME: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

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3 years ago

Read 2 more answers

Need help answering this math problem.

Fed [463]

The second bubble is the answer

4 0

3 years ago

Name three decimals between 0.55 and 0.56

marshall27 [118]

Some decimals between 0.55 and 0.56 are 0.551, 0.552, 0.553, 0.554, 0.556, 0.557, 0.558, 0.559.

i hope this is helpful

5 0

4 years ago

Consider the function f(x) = x² + 10x + 25 for x ≥ -5.<br> What is the value of f-¹(x) when x = 4?

prisoha [69]

Answer:

-3

Step-by-step explanation:

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2 years ago

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected

Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let $x_1$ , $x_2$ , $x_3$ and $x_4$ be four given positive integers and let $x_1+x_2+x_3+x_4= N$ .

A random variable X is said to have hypergeometric distribution with parameter $x_1$ , $x_2$ , $x_3$ , $x_4$ and n.

The probability mass function

$f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }$

Here $a_1+a_2+a_3+a_4=n$

${\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}$

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

$x_1$ =2, $x_2$ =3, $x_3$ =5 and $x_4$ =2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

$a_1$ =1, $a_2$ =1, $a_3$ =1 , $a_4$ =1, n=4

The required probability is

$=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

$P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)$

$=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$ $+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}$

$=\frac{6+12+6}{495}$

$=\frac{8}{165}$

=0.04848

The probability that all nationalities except Italian are represent is 0.04848.

6 0

3 years ago