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babymother [125]
3 years ago
15

please solve the answer and show your work so a 747 boejing jet flying from new york city to hong kong can carry about 160 passe

ngers. the plane travels at an average speed of 715 mph. the distance from new york to hong kong is 8,054 miles. how long is the flight from new york to hong kong? please choose the equation that would solve this problem is it a. 715 divided by 8,054 or b. 8,054 x 715 or c. 8,054 divided by 715 or d. 715 x 160
Mathematics
2 answers:
Igoryamba3 years ago
8 0

Answer:

C, 8054 divided by 715

Step-by-step explanation:

Given: A 747 boeing jet is flying from new york city to hong kong,  the plane travels at an average speed of 715 mph. the distance from new york to hong kong is 8,054 miles.

To Find: How long is the flight from new york to hongkong.

Solution:

Average Speed of boeing during the flight = 715\text{mph}

Distance between newyork city and hongkong = 8054 \text{miles}

Now,

we know that

\text{distance}=\text{speed}\times\text{time}

so,

\text{time}=\frac{\text{distance}}{\text{speed}}

on putting values in the equation

\text{time}=\frac{8054}{715}

                          \text{time}=11.26\text{hours}

Hence the duration of flight between newyork and hongkong is 11.26 hours, the correct answer will be option C. 8054 divided by 715

dybincka [34]3 years ago
6 0
B is correct. 8054 ÷ 715 is
11.2643356643
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X would be 180 - 90 - 35 = 55 degrees

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Simplify:

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Subtract 50 from both sides:

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The store currently has seven rackets of each version. What is the probability that all of the next ten customers who want this
Mila [183]

The probability that all of the next ten customers who want this racket can get the version they want from current stock is 0.821

<h3>How to solve?</h3>

Given: currently has seven rackets of each version.

Then the probability that the next ten customers get the racket they want is P(3≤X≤7)

<h3>Why P(3≤X≤7)?</h3>

Note that If less than 3 customers want the oversize, then more than 7 want the midsize and someone's going to miss out.

X ~ Binomial (n = 10, p = 0.6)

P(3≤X≤7)  = P(X≤7) - P(X≤2)

From Binomial Table:

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               = 0.821

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6 0
2 years ago
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d
aliya0001 [1]

The Lagrangian

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)

has critical points where the first derivatives vanish:

L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}

L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}

L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}

L_\lambda=x^4+y^4+z^4-13=0

We can't have x=y=z=0, since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of \pm\sqrt[4]{13}. For example, if y=z=0, then x^4=13\implies x=\pm\sqrt[4]{13}.

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find \lambda=-\frac1{\sqrt{26}} (taking the negative root because x^2,y^2,z^2 must be non-negative), and we can immediately find the critical points from there. For example, if z=0, then x^4+y^4=13. If both x,y are non-zero, then x^2=y^2=-\frac1{2\lambda}, and

xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}

\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}

and for either choice of x, we can independently choose from y=\pm\sqrt[4]{\frac{13}2}.

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then x^2=y^2=z^2=-\frac1{2\lambda}. We have

xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}

\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}

and similary y,z have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate f at each critical point; you should end up with a maximum value of \sqrt{39} and a minimum value of \sqrt{13} (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

(\sqrt[4]{13},0,0)

(-\sqrt[4]{13},0,0)

(0,\sqrt[4]{13},0)

(0,-\sqrt[4]{13},0)

(0,0,\sqrt[4]{13})

(0,0,-\sqrt[4]{13})

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\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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