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2 years ago

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A total of 351 tickets were sold for the school play. They were either adult tickets or student tickets. The number of students

tickets sold was two times the number of adult tickets sold. How many adult tickets were sold?

1 answer:

Svetlanka [38]2 years ago

3 0

Hi the answer is 117 because 117 times 3 equals 351 , in this type of problems you just have to find a number that when is duplicate is going to give you the answer , so if you add 117 and 117 , 117 , you will find out that your answer is 351.

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According to the fundamental theorem of algebra, how many zeros does the polynomial below have?

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EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

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The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

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$L_z=0$ tells us $\lambda=9$ , so that

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3 years ago

The number of nails of a given length is normally distributed with a mean length of 5 in and a standard deviation of 0.03 find t

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Answer:

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Step-by-step explanation:

Let's first determine the z-score by using the formula:

$z=\frac{x-\mu}{\sigma}$

Where x=5.03, μ=5, σ=0.03 and so:

$z=\frac{5.03-5}{0.03}=1$

We can now use a z-score table to determine that 1.00 is synonymous with 0.8413 meaning 84.13%. This means that 84.13% of the nails are less than or equal to 5.03 inches long. Therefore, since we know 100% = 1 in decimals we can determine that the length greater than 5.03 inches is 1-0.8413=0.1587. This means that 15.87% of the 120 nails are greater than 5.03 inches long.

Finally, we can determine how many nails are greater than 5.03 by determining what is 15.87% of 120 which is:

$120 \times 0.1587 \approx 19$

And so 19 out of the 120 nails are greater than 5.03 inches long.

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3 years ago

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2 years ago

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