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Sloan [31]
3 years ago
7

Please help :(( look at the picture! 8 points !

Mathematics
1 answer:
pochemuha3 years ago
7 0

Answer:

the picture isnt loading can you type the question instead?

Step-by-step explanation:

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URGENT PLEASE HELP PLEASE
mafiozo [28]

Answer: I believe the answer is A.

Explanation: hope this helps

7 0
3 years ago
What is the period of the function shown in the graph?
denpristay [2]

the period for one wave is 8π

6 0
3 years ago
The lengths of pregnancy terms for a particular species of mammal are nearly normally distributed about a mean pregnancy length
snow_lady [41]

Answer: About 99.74% of births would be expected to occur within 24 days of the mean pregnancy length.

Step-by-step explanation:

Complete question is attached below.

Given: The lengths of pregnancy terms for a particular species of mammal are nearly normally distributed about a mean pregnancy length with a standard deviation of 8 days.

i.e. \sigma= 8

let X denotes the random variable that represents the lengths of pregnancy.

The probability of births would be expected to occur within 24 days of the mean pregnancy​ length:

P(\mu-24

= 2(0.9987)-1\ \ \ [\text{ By z-table}]\\\\=0.9974

=99.74%

Hence, about 99.74% of births would be expected to occur within 24 days of the mean pregnancy length.

4 0
3 years ago
Which statement about the graph of these two functions is true?
Anarel [89]
B is true.
a is false because the -4 will cause a downward slant
c is false because f(x) intersects x axis to the right of 0
3 0
3 years ago
Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for
USPshnik [31]

x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}

\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}

6 0
4 years ago
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