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blagie [28]
4 years ago
8

What the answers for this i dont get it

Mathematics
1 answer:
daser333 [38]4 years ago
4 0
It depends on what the question is asking you to find.
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The equation T squared = A cubed shows the relationship between a planet's orbital period, T, and the planet's mean distance fro
Dafna11 [192]

Answer:

Ty = √k  * Tx

The orbital period of Y has to be be multiplied by √k .   or . k∧1/2

Step-by-step explanation:

The general equation:

T² = A³

is for the case of planet X

Tₓ² = Aₓ³

In the case of planet Y

Ty² = k * Aₓ³ .     and   Aₓ³ = Tₓ²

By substitution:

Ty² =  k *Tₓ²

Ty = √k  * Tx

3 0
3 years ago
The sky ride at an amusement park spans 2,715 feet. Over the course of the day,
olga_2 [115]

Answer:

C. 19,005 feet

Step-by-step explanation:

\begin{array}{ccc} \rm 1 \:   time \: ride \implies \: 2175 \: feet\\ \rm 7 \: times \: ride \implies 2175 \times 7 = 19005 \: feet\end{array}

hence,

our answer is C)

7 0
3 years ago
Read 2 more answers
My friend needs help on this problem, plz help, thank you
Dafna1 [17]

Answer:

3 = 95, 4 = 85, if a and b are parallel

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
34. Find each of the following probabilities when n indepen- dent Bernoulli trials are carried out with probability of success p
mr Goodwill [35]

Answer:

A.) (1 - p)^n

B.) 1 - (1 - p)^n

C.) (1 - p)^n + np*(1-p)^(n-1)

D.) 1 - (1 - p)^n - np*(1-p)^(n-1)

Step-by-step explanation:

General form of a binomial probability :

P(x = x) = nCx * p^x * q^(n-x)

q = 1 - p ; n = number of trials ; x = number of successes ; p = probability of success

A.) probability of no successes ;

P(x = 0) = nC0 * p^0 * (1 - p)^(n-0)

P(x = 0) = 1 * 1 * (1 - p)^n

P(x = 0) = (1 - p)^n

Probability of atleast one success = 1 - P(no success)

P(x ≥ 1) = 1 - P(x = 0)

P(x = 0) = (1 - p)^n

P(x ≥ 1) = 1 - P(x = 0) = 1 - (1 - p)^n

Probability of at most one success

P(x ≤ 1) = p(x = 0) + p(x = 1)

P(x = 0) = (1 - p)^n

P(x = 1) = nC1 * p^1 * (1 - p)^(n-1)

P(x = 1) = n * p * (1 - p)^(n-1) = np*(1-p)^(n-1)

P(x ≤ 1) = (1 - p)^n + np*(1-p)^(n-1)

Probability of atleast two successes:

(1 - probability of at most 2 successes)

P(x ≥ 2) = 1 - P(x ≤ 1)

P(x ≥ 2) = 1 - (p(x = 0) + p(x = 1))

P(x ≥ 2) = 1 - p(x = 0) - p(x = 1))

P(x ≥ 2) = 1 - (1 - p)^n - np*(1-p)^(n-1)

6 0
3 years ago
Need help on this one
malfutka [58]

Answer:

3

Step-by-step explanation:

To simplify this expression, we will simply work with the numerator, then the denominator, and then the fraction as a whole, following PEMDAS.

Numerator:  12 - 6 + 5 * (-3)^2

= 12 - 6 + 5 * (9)

= 12 - 6 + 45

= 6 + 45

= 51

Denominator:  17

So our simplified fraction is as follows:

51/17

Which can be reduced to the whole number, 3.

Cheers.

3 0
3 years ago
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