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4 years ago

What is the GCF of 30b^2, 80b^2, and 20b^3?

Mathematics

1 answer:

Nana76 [90]4 years ago

5 0

$30b^2, 80b^2, \ and \ 20b^3$

$30b^2 = 10b^2(3)$

$80b^2 = 10b^2(8)$

$20b^3 = 10b^2(2b)$

$\text {GCF = } 10b^2$

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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.

3 0

4 years ago

the tickets are 90 dollars in all adults 7 kids 12 kids are 2/3 cost of adults tickets. What is the cost of the adult tickets

svetoff [14.1K]

I hope it helps. .........

3 0

3 years ago

Mars2501 [29]

Answer:

$186.86

Step-by-step explanation:

4 eggs cost:

1 dollar per dozen (12 eggs), so by unitary method (ratios), we can find cost of 4 eggs. Shown below:

$\frac{1}{12}=\frac{x}{4}\\12x=4\\x=0.33$

So, 4 eggs cost $0.33 approx

12 lb. flour cost:

2.50 dollar per pound, so 12 pounds would simply cost:

2.5 * 12 = $30

14 lb of sugar cost:

4.50 dollar per pound means 14 pounds would cost:

4.5 * 14 = $63

2 ready-made edible flowers cost:

5 cents for each flower means 2 flowers would cost:

5 cent = 0.05 dollars

0.05 * 2 = $0.10

We add all up to find cost of 1 cake. Shown below:

Cost of 1 cake = $0.33 + $30 + $63 + $0.10 = $93.43

Cost of 2 cakes = $93.43 * 2 = $186.86

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3 years ago

Tannin and Dax bought a new video game for $48 to use the coupon for 25% off how much would the game cost if they paid full pric

olchik [2.2K]

Answer:

$64

Step-by-step explanation:

After using a 25% discount coupon, they paid 48 for the video game.

It means, which number, decreased by 25%, would give us 48?

First, 25% to decimal is:

25/100 = 0.25

Now, if we let the original price be "x", we can say:

x decreased by 0.25 of x would give us 48

Translating this to algebraic equation gives us:

x - 0.25x = 48

Now, we solve this:

$x - 0.25x = 48\\0.75x=48\\x=\frac{48}{0.75}\\x=64$

If they paid full price, the video game would cost $64

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3 years ago

What are the solutions to the system?

artcher [175]

Answer:

Step-by-step explanation:

I just plugged each variable into the equation. A is the only one that works.

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3 years ago