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2 years ago

Combine the following expressions 2√18x^3 - 3√8x^3 A. -√x B. 0 C.√x

Mathematics

1 answer:

rosijanka [135]2 years ago

8 0

This is what I have but it doesn't match with your answers

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Name two ways you might see decimals used outside school

Oxana [17]

Answer:

in money or weighing things.

Step-by-step explanation:

4 0

3 years ago

A public health organization reports that 40%of baby boys 6-8 months old in the United

Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502$

Hence:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675$

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106$

$P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016$

$P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001$

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

4 0

1 year ago

A Ferris wheel at the local fair has a diameter of 36 feet and a midline of 15 feet. This Ferris wheel makes one revolution ever

AVprozaik [17]

Answer:

Hold on let me seee!!!!!!!

5 0

3 years ago

Tank 1 initially contains 50 gals of water with 10 oz of salt in it, while tank 2 initially contains 20 gals of water with 15 oz

Naddik [55]

$\dfrac{\mathrm dx_1}{\mathrm dt}=\dfrac{2\text{ oz}}{1\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}-\dfrac{x_1(t)\text{ oz}}{50\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}$
$\dfrac{\mathrm dx_2}{\mathrm dt}=\dfrac{x_1(t)\text{ oz}}{50\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}-\dfrac{x_2(t)\text{ oz}}{20\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}$

$\implies\begin{cases}\dfrac{\mathrm dx_1}{\mathrm dt}=10-\dfrac1{10}x_1\\\\\dfrac{\mathrm dx_2}{\mathrm dt}=\dfrac1{10}x_1-\dfrac14x_2\\\\x_1(0)=10\\\\x_2(0)=15\end{cases}$

6 0

3 years ago

M=x^3+y^2-6(x-y)-2021

bagirrra123 [75]

Answer:

Step-by-step explanation:

m=x^3+y^2-6(x-y)-2021

distribute: 6x-6y

now you have m=x^3 + y^2 - 6x - 6y - 2021

7 0

2 years ago

Read 2 more answers