How many ways can a person toss a coin 13 times so that the number of heads is between 7 and 10 inclusive?

Mathematics

1 answer:

maks197457 [2]2 years ago

3 0

Answer:

4004

Step-by-step explanation:

The number of combinations of choosing r objects from a group of n is nCr.

So the number of ways to get 7, 8, 9, or 10 heads from 13 coin tosses is:

₁₃C₇ + ₁₃C₈ + ₁₃C₉ + ₁₃C₁₀

= 1716 + 1287 + 715 + 286

= 4004

You might be interested in

2 members of the 5-member math study group will present their project to the class. How many combinations of 2 members are possi

klio [65]

Answer:

Step-by-step explanation:

(,)=?

(,)=(5,2)

=5!(2!(5−2)!)

=5!2!×3!

= 10

4 0

2 years ago

Which function is represented by the graph? f(x) = −2|x| + 1 f(x) = |x| + 1 f(x) = -2|x + 1| f(x) = |x + 1|

nevsk [136]

The answer is A. F(x)= -2 |x| + 1

7 0

2 years ago

Read 2 more answers

Write .125 as a fraction reduced to lowest terms.

VMariaS [17]

Answer:

1/8

Step-by-step explanation:

Hope this helps you! <33

7 0

2 years ago

Read 2 more answers

The equation 12x² + 4kx + 3 = 0 has real and equal roots, if

sleet_krkn [62]

Answer:

$k=\pm3$

Step-by-step explanation:

[...] if you can write the LHS as a perfect square, or if you can't spot a factorization of it right away, if and only if the discriminant $\Delta = b^2-4ac$ (or, if b is an even number, 1/4 of it) is zero.

Stare at it for a while. First term is $3(2x)^2$ , third term is $3(1)^2$ , we are missing a double product, but we can play with k. For the LHS to be $3(2x\pm1)^2 = 3(4x^2\pm4x+1) = 12x^2\pm12x+3$ you just need $4k= \pm12 \rightarrow k=\pm3$ .