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3 years ago

The area of a square piece of cardboard is 81 square inches. What is the length of the cardboard

Mathematics

1 answer:

vredina [299]3 years ago

6 0

9 because the square root of 81 is 9

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3 years ago

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3 years ago

What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C

butalik [34]

ANSWER

$10C_5(2)^{5}( - 3)^5$

EXPLANATION

The given binomial expansion is:

${(2x - 3y)}^{10}$

Compare this to

${(a + b)}^{n}$

we have a=2x , b=-3y and n=10

We want to find the coefficient of the term

${x}^{5} {y}^{5}$

This implies that, r=5.

The terms in the expansion can be obtained using

$T_{r+1}=nC_ra^{n-r}b^r$

We substitute the given values to obtain;

$T_{5+1}=10C_5(2x)^{10-5}( - 3y)^5$

$T_{6}=10C_5(2x)^{5}(3y)^5$

$T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5} {y}^{5}$

Hence the coefficient is;

$10C_5(2)^{5}( - 3)^5$

5 0

3 years ago

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According to the Department of Transportation, 27% of domestic flights were delayed in the last year at JFK airport. Five flight

Mars2501 [29]

Answer:

The probability that all the five flights are delayed is 0.2073.

Step-by-step explanation:

Let X = number of domestic flights delayed at JFK airport.

The probability of a domestic flight being delayed at the JFK airport is, P (X) = p = 0.27.

A random sample of n = 5 flights are selected at JFK airport.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={5\choose x}0.27^{x}(1-0.27)^{5-x};\ x=0,1,2...$

Compute the probability that all the five flights are delayed as follows:

$P(X=5)={5\choose 5}0.27^{5}(1-0.27)^{5-5}=1\times 1\times 0.207307=0.2073$

Thus, the probability that all the five flights are delayed is 0.2073.

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3 years ago