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Nataliya [291]
3 years ago
13

Which set of side lengths is a pythagorean triple

Mathematics
1 answer:
Neko [114]3 years ago
4 0
Several examples of side lengths that are Pythagorean triples are the following with the corresponding side lengths A, B, C:

(5, 12, 13), (7, 24, 25), (3, 4, 5)

-E :)


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Is 3/4 equivalent to 27/36
egoroff_w [7]
Yes because you can multiply 3/4 by 9/9 and you’ll get 27/36 :)
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Find ordered pairs: choose two of the answers below
Anarel [89]

Answer:

b -3,3 c 6,9 hope this helps

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It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey
mote1985 [20]

Answer:

P(X\geq 8)=0.0043\\\\

It's more likely that  all of the residents surveyed will have adequate earthquake supplies since it has a probability of 98.02% which is very close to 100%.

Step-by-step explanation:

-This is a binomial probability problem with the function:

P(X=x)={n\choose x}p^x(1-p)^{n-x}

-Given p=0.3, n=11, the is calculated as:

P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)+P(X=11)\\\\={11\choose 8}0.3^8(0.7)^3+{11\choose 9}0.3^9(0.7)^2+{11\choose 10}0.3^{10}(0.7)^1+{11\choose 11}0.3^{11}(0.7)^0\\\\=0.0037+0.0005+0.00005+0.000002\\\\=0.0043

Hence, the probability that at least 8 have adequate supplies 0.0043

#The probability that non has adequate supplies is calculated as;

P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= 0)={11\choose 0}0.3^{0}(0.7)^{11}\\\\=0.0198

#The probability that all have adequate supplies is calculated as:

P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= All)=1-{11\choose 0}0.3^{0}(0.7)^{11}\\\\=1-0.0198\\\\=0.9802

Hence, it's more likely that  all of the residents surveyed will have adequate earthquake supplies since P(All)>P(None)\ \ and that this probability is 0.9802 or 98.02% a figure close to 1

7 0
3 years ago
Suppose that a function​ f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence
Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

\lim_{x \to c} \frac{x^2-cx}{(x-c)}

We can simplify it

\lim_{x \to c} \frac{x(x-c)}{(x-c)}

=\lim_{x \to c} x

=c

so, we can see that limit exist and it's value defined

case-2:

\lim_{x \to c} \frac{1}{(x-c)}

Left limit is

\lim_{x \to c-} \frac{1}{(x-c)}

=-\infty

Right Limit is

\lim_{x \to c+} \frac{1}{(x-c)}

=+\infty

so, we can see that left limit is not equal to right limit

so, limit does not exist

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3 years ago
Do anybody want to talk
MrRa [10]

Answer:Nooooooope

Step-by-step explanation:

Feeling sleepy...

Good night dear..

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