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Aneli [31]
2 years ago
8

What is the mean of the data set? 33, 29, 18, 24

Mathematics
2 answers:
kodGreya [7K]2 years ago
6 0

Answer:

The mean would be 26.

Step-by-step explanation:

NISA [10]2 years ago
4 0

Answer:

26

Step-by-step explanation:

Here's the formula to find the mean/average.

Add up all numbers.

33+29+18+24=104

Then, divide the result by how many numbers there were. In this case, 4 numbers.

104/4=26

Thus, the answer is 26.

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Evaluate 3 to the 2nd power + (8-2)•4-6/3
erma4kov [3.2K]
3^{2}+(8-2)×4-6/3

Always do what's in parentheses first:

3^{2}+6×4-6/3

Then do multiplication and division:

9+24-2

Then addition and subtraction last:

9+24-2=31
8 0
3 years ago
This is my question. If Andy drove 840 miles in 12 hours. How far could he drive in hours? Help me?
stellarik [79]

Answer:

840mi. - 12 hrs

  x       -  3 hrs

This is a simple cross multiplying problem.

840 _ 12

 x   --  3    ⇒12x=3(840) ⇒ 4x=280 ⇒ x=70 miles

:D

Step-by-step explanation:

7 0
3 years ago
How do you find the next three terms in the geometric sequence 9, -3, 1, , ... ?
levacccp [35]
Each term is being divided by -3, so the next three terms would be:

-1/3, 1/9, -1/27
7 0
2 years ago
Given: 315/1575. Write the prime fractorization of the numerator and the denominator, then write the fraction in lowest terms.
Ne4ueva [31]
The prime factorization of the numerator, 315, is 5 x 3 x 3 x 7. That of the denominator, 1575 is 5 x 5 x 3 x 3 x 7. To convert the fraction to its lowest form, cancel all the factors that appear to both the numerator and denominator. We will be left with 1/5. 
3 0
2 years ago
Reduced form of 2ab^2-a^2 b^2/5
Ket [755]

Answer:

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

Step-by-step explanation:

Given the expression

2ab^2-\frac{a^2b^2}{5}

\mathrm{Convert\:element\:to\:fraction}:\quad \:2ab^2=\frac{2ab^25}{5}

2ab^2-\frac{a^2b^2}{5}=\frac{2ab^2\cdot \:5}{5}-\frac{a^2b^2}{5}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

                 =\frac{2ab^2\cdot \:5-a^2b^2}{5}

                 =\frac{10ab^2-a^2b^2}{5}

Thus,

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

7 0
2 years ago
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