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3 years ago

Consider the figure. What is JL?

Mathematics

1 answer:

lisov135 [29]3 years ago

3 0

The answer for the above mentioned problem is JL = 12.5

Step by step explanation:

Given:

JM = 8

KM = 6

To Find:

JL = ?

Formula to be used:

$KM^2$ = JM x ML

In order to find " JL" we must first find "ML",

$KM^2$ = JM x ML

$6^2$ = 8 x ML

36 = 8 x ML

36/8 = ML

ML = 4.5

Now JL = 4.5+8

= 12.5

Thus the value of JL = 12.5

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2 years ago

(x +y)^5 Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

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2 years ago

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following p

aev [14]

Answer:

P(X < 3) = 0.7443

Step-by-step explanation:

We are given that the random variable X has a binomial distribution with the given probability of obtaining a success. Also, given n = 6, p = 0.3.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 6

r = number of success = less than 3

p = probability of success which in our question is 0.3.

LET X = a random variable

So, it means X ~ $Binom(n=6, p=0.3)$

Now, Probability that X is less than 3 = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{6}{0}0.3^{0} (1-0.3)^{6-0}+ \binom{6}{1}0.3^{1} (1-0.3)^{6-1}+ \binom{6}{2}0.3^{2} (1-0.3)^{6-2}$

= $1 \times 1 \times 0.7^{6} +6 \times 0.3^{1} \times 0.7^{5} +15 \times 0.3^{2} \times 0.7^{4}$

= 0.11765 + 0.30253 + 0.32414 = 0.7443

Therefore, P(X < 3) = 0.7443.

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3 years ago