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snow_lady [41]
3 years ago
12

How do you find the equation of the axis of symmetry on a quadratic graph

Mathematics
1 answer:
andrew-mc [135]3 years ago
5 0

It is the line of symmetry of a parabola and divides a parabola into two equal halves that are reflections of each other about the line of symmetry.


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Un árbol de 4.8 m de altura proyecta una sombra de 6.3 m. Encuentra la medida del ángulo "x" de elevación del sol.
Semenov [28]

Answer:

37 degrees

Step-by-step explanation:

Tan x = 4.8/6.3

Tan x = 0.76

Tan^-1 (0.76) = 37 degrees

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The monthly worldwide average number of airplane crashes of commercial airlines is 2.2. What is the probability that there will
Makovka662 [10]

Answer:

(a) more than 2 such accidents in the next month \approx 0.3773

(b) more than 4 such accidents in the next 2 months \approx 0.44882

(c) more than 5 such accidents in the next 3 more than 5 such accidents in the next 3 months \approx 0.64533

Step-by-step explanation:

Let  N be the  Random variable that marks the number of crashes in certain month.

Now let us use Poisson distribution since we are given with average number of crashes that is N \sim Pois(2.2)

(A)  more than 2 such accidents in the next month

Probability(more than 2 such accidents in the next month)=P(N>2)

P(N>2)=1-P(N=0)-P(N=1)-P(N=2)

=>1-e^-{2.2}-2.2e^{-2.2}-\frac{2.2^2}{2!}e^{2.2}

=> \approx 0.3773

B) more than 4 such accidents in the next 2 months

since the average number of crashes in 1 month is 2.2, the average number of crashes in two months is 4.4. hence, if we say that N_1 is the number of crashes  in 2 months, we have that N\simPois(4.4)

Thus,

Probability(more than 4 such accidents in the next 2 months)=P(N_1>4)

=1-P(N_1=0)-P(N_1=1)-P(N_1=2)=P(N_1=3)-P(N_1=4)

1-\sum_{k=1}^{4} \frac{4.4^{k}}{k !} e^{-4.4}

=> \approx 0.44882

C)  more than 5 such accidents in the next 3 more than 5 such accidents in the next 3 months

If we say that N_2 marks the number of crashes in the next 3 months , using the same argument as in (a) we have that  a N\simPois(6.6)

Hence

P(N_2>5)=1-\sum_{k=0}^{5} \frac{6.6^{k}}{k !} e^{-6.6}

=>\approx 0.64533

6 0
3 years ago
Someone help me with this and I’ll mark u as brilliant
DedPeter [7]

Answer:

B)

Step-by-step explanation:

3 0
3 years ago
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