What will be the value of the 22nd coefficient of a binomial raised to the power 25

Mathematics

1 answer:

Alona [7]2 years ago

4 0

Step-by-step answer:

We are looking at the coefficient of the 22nd term of (x+y)^25.

Following the sequence, first term is x^0y^25, second term is x^1y^24, third term is x^2y^23...and so on, 22nd term is x^21y^4.

The twenty-second term of (x+y)^25 is given by the binomial theorem as

( 25!/(21!4!) ) x^21*y^4

=25*24*23*22/4! x^21y^4

= 12650 x^21 y^4

The coefficient required is therefore 12650, for a binomial with unit valued coefficients.

For other binomials, substitute the values for x and y and expand accordingly.

Question would have been more clearly stated if the actual binomial was given, as commented above.

You might be interested in

What is the line of semmetry for the parabola whose equation is y =x^2 +10x=25

frozen [14]

Answer:

x = -5

Step-by-step explanation:

The equation y = x^2+10x+25 can be factored to (x+5)^2

when (x+5)^2 = 0

x = -5

meaning the equation has a line of symmetry at x = -5

5 0

2 years ago

Freidy is practicing for the marathon. It takes him 1.75 hours to run 12.25 miles. At this rate, how long will it take you to co

Flauer [41]

Answer:

It would take him about 3.7 hours to run 26.2 miles because he can run 7 miles per hour and 26.2/7 is 3.7....

Step-by-step explanation:

hope this helps :)

5 0

3 years ago

Read 2 more answers

A simple random sample of size n=250 individuals who are currently employed is

Likurg_2 [28]

Answer:

$0.104 - 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.054$

$0.104 + 2.58\sqrt{\frac{0.104(1-0.104)}{250}}=0.154$

The 99% confidence interval would be given by (0.054;0.154) . So we are confident at 99% that the true proportion of people that they did work at home at least once per week is between 0.054 and 0.154

Step-by-step explanation:

For this case we can estimate the population proportion of people that they did work at home at least once per week with this formula:

$\hat p = \frac{X}{n}= \frac{26}{250}= 0.104$

We need to find the critical value using the normal standard distribution the z distribution. Since our condifence interval is at 99%, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: