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3 years ago

10

What postulate is this? AAS ASA SAS AAA

2 answers:

seropon [69]3 years ago

5 0

SAS: side, angle, side

nata0808 [166]3 years ago

4 0

The answer is ASA(side,angle,side)

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Is it true or false that a bbinomial can have a degree of zero?

Mumz [18]

No, because any real valued term with degree 0 would be a constant. All constants in an expression can be combined into single term. Therefore if the expression were of degree 0, all the terms would be constants and could be combined into a single term making the expression a monomial.

6 0

3 years ago

Read 2 more answers

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

Which point has the coordinate –0.4?

Sphinxa [80]

The best and the most correct answer among the choices provided by the question is the fourth choice. K has a coordinate of -0.4. I hope my answer has come to your help. God bless and have a nice day ahead! Feel free to ask more questions.

8 0

2 years ago

Which describes these 2 figures

Sophie [7]

The answer for this is A

6 0

2 years ago

Read 2 more answers

What are the domain and range of f (x) = log (x minus 1) 2?.

statuscvo [17]

You can use the definition of logarithm and the fact that a positive number raised to any power will always stay bigger than 0.

The domain of the given function is {x | x > 1 and a real number }

The range of the given function is $\mathbb R$ (set of real numbers)

<h3>What is the definition of logarithm?</h3>

If a is raised to power b is resulted as c, then we can rewrite it that b equals to the logarithm of c with base a.

Or, symbolically:

$a^b = c \implies b = log_a(c)$

Since c was the result of a raised to power b, thus, if a was a positive number, then a raised to any power won't go less or equal to zero, thus making c > 0

<h3>How to use this definition to find the domain and range of given function?</h3>

Since log(x-1) is with base 10 (when base of log isn't specified, it is assumed to be with base 10) (when log is written ln, it is log with base e =2.71828.... ) thus, we have a = 10 > 0 thus the input x-1 > 0 too.

Or we have:

x > 1 as the restriction.

Thus domain of the given function is {x | x > 1 and a real number }

Now from domain, we have:

$x > 1\\
x-1 > 0\\
log(x-1) > -\infty\\
log(x-1) + 2 > -\infty\\
f(x) > -\infty$ (log(x-1) > -infinity since log(0) on right side have arbitrary negatively large value which is denoted by -infinity)

Thus, range of given function is whole real number set $\mathbb R$ (since all finite real numbers are bigger than negative infinity)

Thus, the domain of the given function is {x | x > 1 and a real number }

The range of the given function is $\mathbb R$ (set of real numbers

Learn more about domain and range here:

brainly.com/question/12208715

8 0

2 years ago