WILL MARK BRAINLIEST<br>solve for x: m(2m-x)=3(x+6) if m≠3

Simplify the expression below, I really just need the steps I have the answer (also can someone tell me how to edit points on th

Sloan [31]

Answer: $3x^2y\sqrt[3]{y}\\\\$

Work Shown:

$\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\$

Explanation:

As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.

4 0

2 years ago

Chef Daniel had 431 ounces of chocolate frosting in his refrigerator. It takes 8 ounces of frosting to frost one cupcake. How ma

Alex

Answer:

53 cupcakes

Step-by-step explanation:

Take the amount of frosting and divide by the amount needed per cupcake

431/8

53.875

We round down since we cannot frost part of a cupcake

53 cupcakes

7 0

3 years ago

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Joe flips the disc shown below 15 times:

vfiekz [6]

Whatever the number of times you flip the disc, as long as the disc has uniformly distributed mass throughout the body, then

The probability will always be 50%

7 0

2 years ago

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The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.

devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0

3 years ago

Please help me. I am struggling. Question is in image.

Anni [7]

Step-by-step explanation:

it is 6 and 7 so that's because Tom Holland is baby daddy

8 0

1 year ago

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WILL MARK BRAINLIESTsolve for x: m(2m-x)=3(x+6) if m≠3