All categories

3 years ago

Harriet rolls a number cube what is the probability that the number cube will land on three or four

Mathematics

1 answer:

makvit [3.9K]3 years ago

6 0

Her chances of getting that should be 2/6

You might be interested in

Which choice gives the radius of the circle represented by the equation? x squared y squared 225

myrzilka [38]

X^2 + y^2 = 225

The radius is 15.
ANswer is C

8 0

3 years ago

Help pls due in 5 minutes

coldgirl [10]

Answer:

Step-by-step explanation:

$2(\frac{7\times 6}{2})+3(13.1\times 7)$

ANSWER IS D

4 0

3 years ago

I REALLY NEED HELP PLEASE The measures of three line segments are given in each set Which set of line segments cannot form a tri

aksik [14]

Answer:

The Answer is D

Step-by-step explanation:

K12

4 0

3 years ago

Hiw many solutions does this have

Snezhnost [94]

Answer:

one solution ,also hi

3 0

3 years ago

Read 2 more answers

Can someone tell me how

gulaghasi [49]

the way I get the subsequent term, nevermind the exponents, the exponents part is easy, since one is decreasing and another is increasing, but the coefficient, to get it, what I usually do is.

multiply the current coefficient by the exponent of the first-term, and divide that by the exponent of the second-term + 1.

so if my current expanded term is say 7a³b⁴, to get the next coefficient, what I do is (7*3)/5 <----- notice, current coefficient times 3 divided by 4+1.

anyhow, with that out of the way, lemme proceed in this one.

$\bf ~~~~~~~~\textit{binomial theorem expansion} \\\\ \qquad \qquad (1+ax)^n\implies \begin{array}{llll} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(1)^n(ax)^0\\\\ 2&+\frac{(1)(n)}{1}\to n&(1)^{n-1}(ax)^1\\\\ 3&+\frac{n\cdot (n-1)}{2}&(1)^{n-2}(ax)^2 \end{array}$

so, following that to get the next coefficient, we get those equivalents as you see there for the 2nd and 3rd terms.

so then, we know that the expanded 2nd term is 24x therefore

$\bf n(1)^{n-1}(ax)1 = 24x\implies n(1)(ax)=24x\implies nax=24x\implies n=\cfrac{24}{a}$

we also know that the expanded 3rd term is 240x², therefore we can say that

$\bf \cfrac{n(n-1)}{2}~~(1)^{n-2}(ax)^2 = 240x^2\implies \cfrac{n(n-1)}{2}(1)(a^2x^2) = 240x^2 \\\\\\ \cfrac{(n^2-n)(a^2x^2)}{2}=240x^2\implies \cfrac{(n^2-n)(a^2)}{2}=\cfrac{240x^2}{x^2}\implies \cfrac{a^2n^2-a^2n}{2}=240 \\\\\\ a^2n^2-a^2n=480$

but but but, we know what "n" equals to, recall above, so let's do some quick substitution

$\bf a^2n^2-a^2n=480\qquad \boxed{n=\cfrac{24}{a}}\qquad a^2\left( \cfrac{24}{a} \right)^2-a^2\left( \cfrac{24}{a} \right)=480 \\\\\\ a^2\cdot \cfrac{24^2}{a^2}-24a=480\implies 24^2-24a=480\implies 576-24a=480 \\\\\\ -24a=-96\implies a=\cfrac{-96}{-24}\implies \blacktriangleright a = 4\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ n=\cfrac{24}{a}\implies n=\cfrac{24}{4}\implies \blacktriangleright n=6 \blacktriangleleft$

7 0

3 years ago