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3 years ago

An art project has an area of 63.6 square inches. What is the width, in inches, of the project?

Mathematics

1 answer:

masya89 [10]3 years ago

5 0

<em>The width of an art project that has an area of 63.6 square inches is roughly 7.94 inches</em>

<em />

<h2>Explanation:</h2>

An art project has an area of 63.6 square inches, and we are asked to find the width, in inches, of the project. So we can easily assume that the art is a square. The area for any square is given by the following expression:

$A=s^2\\ \\ s:A \ side \ of \ the \ square \\ \\ Here \ the \ side \ (s) \ is \ called \ the \ width, \ so: \\ \\ \\ A=63in^2 \\ \\ s^2=63 \\ \\ Taking \ square \ root: \\ \\ s=\sqrt{63} \\ \\ \s=3\sqrt{7}in \\ \\ s \approx 7.94 in$

<h2>Learn more:</h2>

Area of circles: brainly.com/question/13878095

#LearnWithBrainly

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Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

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3 years ago

4. The length of a rectangle is sixteen inches less

-BARSIC- [3]

Answer:

68in

Step-by-step explanation:

l = 3w-16

l-3w = -16

l+w = 96

-4w = -112

w = 28 in

l = 96-28 = 68in

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2 years ago

Without graphing, tell whether the point (10, 37) is on the graph of y = 3x + 7.

slava [35]

Answer: YES

Plug in the numbers in your head.

y=3(10)+7

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y=37 which is the y intercept

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3 years ago

Suppose an isosceles triangle ABC has A = 45° and b = c = 4. What is the length of a^2?

Mila [183]

Since b = c, the angles B and C are congruent.
180 - 45 = 135
135/2 = =67.5

A = 45
B = 67.5
C = 67.5
a = ?
b = 4
c = 4

Since we know two sides and all three angles, we can use the law of sines to find a.

b/sin B = a/sin A

4/(sin 67.5) = a/(sin 45)

a = 4(sin 45)/(sin 67.5)

a^2 = (4(sin 45)/(sin 67.5))^2

a^2 = 9.37

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jeyben [28]

Answer:

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