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2 years ago

can you find the lengths and area and type the correct code? Please remember to type in all caps with no space

Mathematics

2 answers:

Yakvenalex [24]2 years ago

5 0

hahahahahahahahahahahahahahahahahaha

Fofino [41]2 years ago

4 0

Answer:

EGFB

Step-by-step explanation:

To find the lengths of PQ and RS, we can use the Pythagorean Theorem, becasue both triangles have right angles. Base x height equals the area of the rectangle, and base x height divided by 2 gets us the height of the triangle. Thus:

1. 233 meters

2. 21, 840 square meters

3. 241.87 meters

4. 12, 600 meters

Hope this helps.

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2 years ago

Suppose you toss a fair coin 10 times, let X denote the number of heads. (a) What is the probability that X=5? (b) What is the p

zubka84 [21]

Answer: The required answers are

(a) 0.25, (b) 0.62, (c) 6.

Step-by-step explanation: Given that we toss a fair coin 10 times and X denote the number of heads.

We are to find

(a) the probability that X=5

(b) the probability that X greater or equal than 5

(c) the minimum value of a such that P(X ≤ a) > 0.8.

We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :

$P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.$

(a) The probability of getting 5 heads is given by

$P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.$

(b) The probability of getting 5 or more than 5 heads is

$P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.$

(c) Proceeding as in parts (a) and (b), we see that

if a = 10, then

$P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.$

Therefore, the minimum value of a is 6.

Hence, all the questions are answered.

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4 years ago

What is the binomial expansion of (2x – 3)^5?

Tanya [424]

Answer:

Step-by-step explanation:

(2x + 3)^5 = C(5,0)2x^5*3^0 +

C(5,1)2x^4*3^1 + C(5,2)2x^3*3^2 + C(5,3)2x^2*3^3 + C(5,4)2x^1*3^4 + C(5,5)2x^0*3^5

Recall that

C(n,r) = n! / (n-r)! r!

C(5,0) = 1

C(5,1) = 5

C(5,2) = 10

C(5,3) = 10

C(5,4) = 5

C(5,5) = 1

= 1(2x^5)1 + 5(2x^4)3 + 10(2x^3)3^2 + 10(2x^2)3^3 + 5(2x^1)3^4 + 1(2x^0)3^5

= 2x^5 + 15(2x^4) + 90(2x^3) + 270(2x^2) + 405(2x) +243

= 32x^5 + 15(16x^4) + 90(8x^3) + 270(4x^2) + 810x + 243

= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243

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4 years ago

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3 years ago

can you find the lengths and area and type the correct code? Please remember to type in all caps with no space​

can you find the lengths and area and type the correct code? Please remember to type in all caps with no space