# A high school auditorium seats 110 people. The school play has 106 people in attendance leaving 4 seats empty.because the order in whichThere arethe seats are chosenways that 4 seats can be left empty in the auditorium. This is aimportant.

5773185

Step-by-step explanation:

There are 110 seats

110 ways to choose the first empty seat

Now there are 109 seats

109 ways to choose the next empty seat

Now there are 108 seats

108 ways to choose the next empty seat

Now there are 107 seats

110*109*108*107=138556440

Now the order of the empty seats doesn't matter so we need to divide by 4!

138556440/ 4!

138556440/ 24

5773185

In this mathematics problem, we are asked to determine the number of ways that 4 seats can be left empty in a high school auditorium that seats 110 people. We can use the concept of combinations to solve this.

### Explanation:

In this problem, we are asked to determine the number of ways that 4 seats can be left empty in a high school auditorium that seats 110 people. To solve this, we can use the concept of combinations. The total number of ways to choose 4 seats out of 110 is represented by the combination formula: C(110, 4). To calculate this, we can use the formula: C(n, r) = n! / (r!(n - r)!), where n is the total number of seats and r is the number of seats left empty. Plugging in the values, we have C(110, 4) = 110! / (4!(110 - 4)!).

Using a calculator, we can simplify this expression and calculate the answer.

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## Related Questions

A pump discharges 278 gallons of water in 3.5 minutes. How long would it take to empty a container with 750 gallons? Round up to the nearest minute

It would take 10.5 minutes

A stop sign is in the shape of a regular octagon. Each side measures 12.4 inches and the apothem of the octagon measures 15 inches. A stop sign is shown. What is the area of the stop sign? 93 square inches 186 square inches 744 square inches 1,488 square inches

The area of the stop sign which is an octagon is 744 square inches.

Option C is the correct answer.

We have,

To find the area of the stop sign, we can use the formula for the area of a regular polygon:

Area = (1/2) × perimeter × apothem

The perimeter of the stop sign is 8 × 12.4 = 99.2 inches.

Using the given apothem of 15 inches, we can plug in the values and calculate:

Area = (1/2) × 99.2 × 15

Area = 744 square inches

Therefore,

The area of the stop sign is 744 square inches.

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Step-by-step explanation:

Just did it it correct trust me

What is the product of 4.201 and 5.3? Round to the nearest hundredth

22.2653 or 22.27

Step-by-step explanation:

hope this helps

Step-by-step explanation:  4.201x5.3 = 22.265

Find the derivation of y=sin (2x2 + 3x-4)<br />a. -cos (4x +3) b. -(6x2+3) cos (2x3+3x-4) c. Cos (2x3+3x-4)<br />d. (6x3+3) cos (2x3+3)<br />​

Let y = f(x)

f(x) = sin x

g(x) = (2x^2 + 3x - 4)

Use the chain rule.

We want y' = f' (g(x)) • g'(x).

y' = cos(2x^2 + 3x - 4) • (4x + 3)

Done.

For the characteristic polynomialp(s) =s5+ 2s4+ 24s3+ 48s2−25s−50(a) Use the Routh-Hurwitz Criterion to determine the number of roots ofp(s) in the right-half plane, in the left-half plane, and on thejω-axis.(b) Use Matlab to determine the roots ofp(s), and verify your results in part 2a.

• 1 root in the right half-plane
• 1 conjugate pair on the imaginary axis
• 2 roots in the left half-plane

Step-by-step explanation:

Without using the Routh-Hurwitz criterion at all, you know there is one positive real root. Descartes' rule of signs tells you the number of positive real roots is equal to the number of sign changes in the coefficients (perhaps less a multiple of 2). There is one sign change in + + + + - - , so there is one positive real root.

_____

(a) The Routh array starts as two rows of the polynomial's coefficients, alternate coefficients on each row. For this odd-degree polynomial, the number of coefficients is even, so no zero-padding is necessary at the right end of the second row. That is, we start with ...

The next row is formed from combinations of coefficients in the two rows above. The computation is similar to that of a determinant. By matching the numbers to those in the array, you can see the pattern of the computation.

The next row values are ...

Simplifying, we find this row to be ...

The zero row is a special case that requires we proceed as follows. The row above (identified with s⁴) represents an "auxiliary polynomial":

To continue the process, we replace the zero row by the coefficients of the derivative of this auxiliary polynomial. Proceeding as before, the array now becomes ...

The number of sign changes in the first column (1) tells the number of roots in the right half-plane. The auxiliary polynomial will give us the remaining two pairs of roots:

So, we have determined there to be ...

• 1 root in the right half-plane
• 2 roots on the jω axis
• 2 roots in the left half-plane

__

(b) The original polynomial can be factored as ...

p(s) = (s +2)(s² +25)(s +1)(s -1)

p(s) = (s +2)(s +1)(s -5i)(s +5i)(s -1)

This verifies our result from part (a).

_____

Any row can be multiplied by a convenient factor to simplify the arithmetic. Here, it would be convenient to divide the second row by 2 and the third row by 8.

A zero element (not row) in the first column is replaced by "epsilon" (a small positive number) and the rest of the arithmetic is continued as normal. That row is not counted (it is ignored) when counting sign changes in the first column.

The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ=0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but you believe that the mean nicotine content is actually higher than advertised. To explore this, you test the hypotheses H0:μ=1.5, Ha:μ>1.5 and you obtain a P-value of 0.052. Which of the following is true? A. At the α=0.05 significance level, you have proven that H0 is true. B. This should be viewed as a pilot study and the data suggests that further investigation of the hypotheses will not be fruitful at the α=0.05 significance level. C. There is some evidence against H0, and a study using a larger sample size may be worthwhile. D. You have failed to obtain any evidence for Ha.

Step-by-step explanation:

This is a test of a single population mean since we are dealing with mean.

From the information given,

Null hypothesis is expressed as

H0:μ=1.5

The alternative hypothesis is expressed as

Ha:μ>1.5

This is a right tailed test

The decision rule is to reject the null hypothesis if the significance level is greater than the p value and accept the null hypothesis if the significance level is less than the p value.

p value = 0.052

Significance level, α = 0.05

Since α = 0.05 < p = 0.052, the true statement would be

At the α=0.05 significance level, you have proven that H0 is true. B.