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Anna007 [38]
3 years ago
12

Simplify open parentheses x to the 2 third power close parentheses to the 4 fifths power.

Mathematics
2 answers:
KonstantinChe [14]3 years ago
6 0

Answer:  x^{\frac{8}{15}}

Or x to the power of 8/15.

Step-by-step explanation:

Here, the given expression is,

(x^{\frac{2}{3}})^{\frac{4}{5}}

=x^{\frac{2}{3}\times \frac{4}{5}}      ((a^m)^n=a^{m\times n})

=x^{\frac{8}{15}}

Which is the required simplified of the given expression.

MissTica3 years ago
3 0
X to the 8/15 power. When you have a power raised to another power, you multiply the exponents.
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Two angles in a triangle are 30 degrees and 40 degrees. Find the missing angle measure. Please show your math work step by step
brilliants [131]

Answer:

110

Step-by-step explanation:

Hello There!

If you didn't know the rule about angles in a triangle is that they have a sum of 180

So we can find the measure of the missing angle by subtracting the given angles (in this case 30 and 40) from 180

so the missing angle = 180 - 30 - 40

180 - 30 = 150

150 - 40 = 110

so the missing angle = 110

5 0
3 years ago
How do I solve this math problem 5/6n=10​
GaryK [48]

Answer:

Step-by-step explanation:

5/6n=10​

Cross multiply

5 = 10×6n

5 = 60n

n = 5/60

n = 1/12

7 0
3 years ago
Read 2 more answers
At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample
omeli [17]

Given Information:

Population mean = p  = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

Answer:

Sample mean = μ = 0.60

standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

50*0.60 ≥ 10

30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

50(1 - 0.60) ≥ 10

50(0.40) ≥ 10

20 ≥ 10  (satisfied)

The mean of the sampling distribution will be same as population mean that is

Sample mean = p = μ = 0.60

The standard deviation for this sampling distribution is given by

\sigma = \sqrt{\frac{p(1-p)}{n} }

Where p is the population mean that is proportion of female students and n is the sample size.

\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma =  0.069

Therefore, the standard deviation of the sampling distribution is 0.069.

4 0
3 years ago
The ordered pair (3,9) is a solution to the system of
nadya68 [22]

Answer:

(2,7) is not a solution to the given system of equations.

Step-by-step explanation:

Given system of equation is:

2x + 3 = y

2x + y = 15

To check whether (2,7) is solution to this system or not, we will put x=2 and y=7 in both equations.

Putting x=2 and y=7 in Eqn 1

2(2) + 3 = 7

4 + 3 = 7

7 = 7

Thus the ordered pair satisfies the equation

Putting x=2 and y=7 in Eqn 2

2(2) + 7 = 15

4 + 7 = 15

11 ≠ 15

The ordered pair do not satisfy the second equation.

Hence,

(2,7) is not a solution to the given system of equations.

8 0
3 years ago
3)
Wewaii [24]

Answer:

For covering  1 unit area of the entire playground, the amount of sand required is equal to volume of 3 buckets of sand.

Step-by-step explanation:

Given -

\frac{1}{3} volume of sand in bucket is able to cover \frac{1}{9} area of the entire playground

Thus,

For covering  \frac{1}{9} unit area of the entire playground, the amount of sand required is equal to  \frac{1}{3}  of the total volume of sand in bucket

For covering  1 unit area of the entire playground, the amount of sand required is equal to

\frac{\frac{1}{3} }{\frac{1}{9} } \\\\\frac{1}{3}  * \frac{9}{1} \\\frac{9}{3}\\= 3

For covering  1 unit area of the entire playground, the amount of sand required is equal to volume of 3 buckets of sand.

4 0
3 years ago
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