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Nimfa-mama [501]
2 years ago
5

A local park has a grassy playground that is 3/5 mile long and 1/3 mile wide. What part of a square mile does the playground cov

er? Enter your answer, in simplest form, in the boxes.
?

----mi²

?
Mathematics
1 answer:
adell [148]2 years ago
5 0

Answer:

The playground covers \frac{1}{5} of a square mile


Step-by-step explanation:

The playground is a rectangular ground. The area of a rectangle is given by legth * width, hence,

The area of the playground is length * width

Area = \frac{3}{5}*\frac{1}{3}=\frac{3}{15}=\frac{1}{5} sq. mi


Hence the playground cover \frac{1}{5} of a square mile

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A. Is the sum of two polynomials always a polynomial? Explain
IgorLugansk [536]

Answer:  The answers for both (a) and (b) is YES.


Step-by-step explanation: A polynomial is an algebraic expression containing   two or more algebraic terms, i.e., the sum of several terms that contain different powers of the same variable or variables with real coefficients.

For example, p(x) = 4x²+x+2 is a polynomial in variable 'x'.

(a) Yes, the sum of two polynomials is again a polynomial. For example,

if p(x) = ax² + bx + c  and  q(x) = dx² + ex + f, where, a, b, c, d, e and f are real numbers, then their sum will be

p(x) + q(x) = (a+d)x²+(b+e)x+(c+f), which is again a polynomial in 'x' with real coefficients.

(b) Yes, the difference of two polynomials is again a polynomial. For example,

p(x) - q(x) = (a-d)x²+(b-e)x+(c-f), which is again a polynomial in 'x' with real coefficients.

Thus, the answer is YES.

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2 years ago
Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the fiv
Dahasolnce [82]

Answer and Explanation:

Given : Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029

\sum P(X)=1

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

\mu=\sum xP(x)=2.5

The standard deviation of the random sample is

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\sigma^2=\sum x^2P(x)-\mu^2

\sigma^2=7.436-(2.5)^2

\sigma^2=7.436-6.25

\sigma^2=1.186

\sigma=\sqrt{1.186}

\sigma=1.08

Therefore, The mean is 2.5 and the standard deviation is 1.08.

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