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lutik1710 [3]
3 years ago
10

How do I simplify the expression 2-^1/3 times 2^7/3 A. 64 B. 1/4 C. 4 D. 8

Mathematics
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer:

C.4

Step-by-step explanation:

Since the values are of the same base....

we add the powers..because its multiplication...

;2^(-1/3 + 7/3)

;2^(6/3) = 2^2

;Ans = 2^2 = 4

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Socialization is cultural accumulation.<br><br><br> TrueFalse<br><br> will give brainliest
madam [21]

Answer:

true

Step-by-step explanation:

this is the answer that I think you are looking for.

6 0
2 years ago
Decrease 248 by 30%
Snowcat [4.5K]
Okay this equation really says is what is 30% of 248.

So, lets convert 30% to a fraction, 3/10 which is easier to work with.

All you have to do now is get out a calculator and do 248 *3/10 (or .3) and get 74.4

So subtract 74.4 and get

173.6
8 0
3 years ago
Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t
Gnom [1K]

Answer:

Second class have higher marks and greater spread.

Step-by-step explanation:

First box plot  represents class first. From the first box plot, we get

\text{Minimum value }= 53,Q_1=62,Median=80,Q_3=86,\text{Maximum value }= 89

Range=Maximum-Minimum=89-53=36

IQR=Q_3-Q-1=86-62=24

Second box plot  represents class second. From the second box plot, we get

\text{Minimum value }= 56,Q_1=62,Median=74,Q_3=89,\text{Maximum value }= 96

Range=Maximum-Minimum=96-56=40

IQR=Q_3-Q-1=89-62=27

First class has greater minimum value, first quartile of both classes are same, second class has greater median, first class has greater third quartile and first class has greater maximum value. It means second class have higher marks but class first have less variation.

Second class has greater range and greater inter quartile range. It means data of second class has greater spread.

Therefore, second class have higher marks and greater spread.

3 0
3 years ago
A fence must be built to enclose a rectangular area of 45 comma 000 ftsquared. Fencing material costs $ 3 per foot for the two s
Tatiana [17]

Answer:

150 feet by 300 feet.

Step-by-step explanation:

The fence is to enclose a rectangular area of 45,000 ft squared.

If the dimensions of the rectangle are x and y

Area of a rectangle = xy

  • xy=45000
  • x=\frac{45000}{y}

Perimeter of the Rectangle =2x+2y

Fencing material costs $ 3 per foot for the two sides facing north and south and ​$6 per foot for the other two sides.

  • Cost of Fencing, C=$(6*2x+3*2y)=$(12x+6y)

Substitute x=\frac{45000}{y} into the Cost to get C(y)

C=12x+6y

C(y)=12(\frac{45000}{y})+6y\\C(y)=\frac{540000+6y^2}{y}

The value at which the cost is least expensive is at the minimum point of C(y), when the derivative is zero.

C^{'}(y)=\dfrac{6y^2-540000}{y^2}

\dfrac{6y^2-540000}{y^2}=0\\6y^2-540000=0\\6y^2=540000\\y^2=\frac{540000}{6} =90000\\y=\sqrt{90000}=300

Recall,

x=\frac{45000}{y}=\frac{45000}{300}=150

Since x=150, y=300

The dimensions that will be least expensive to build is 150 feet by 300 feet.

6 0
3 years ago
3 (2x -1) = -2 (x +3)
Misha Larkins [42]

Answer:

x = 3/8 (but the fraction is negative.)

Decimal form if needed.

x = - 0.375

6 0
3 years ago
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