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horrorfan [7]
3 years ago
5

You buy 2.3 pounds of apples for$1.43 per pound. How much doyou spend?​

Mathematics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer:

$3.30 ($3.289)

Step-by-step explanation:

2.3 times 1.43 = 3.289 rounded to 3.30

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Answer:

13

Step-by-step explanation:

Draw a picture!

You have a right triangle with legs of 5 and 12.

By the Pythagorean Theorem, the hypotenuse is √(5²+12²) = 13

3 0
3 years ago
John scored 75, 82, 66, and 78 on his history tests. What score must John get on his next test in order to have a mean score of
Pachacha [2.7K]
[75 + 82 + 66 + 78 + x ] / 5 = 76

301 + x = 380

x = 79
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3 years ago
The scores on a statistics test had a mean of 81 and a standard deviation of 9. One student was absent on the test day, and his
Rama09 [41]

Using the definitions of the mean and of the standard deviation of a data-set, we have that:

  • The mean would increase.
  • The standard deviation would remain constant.

<h3>What are the mean and the standard deviation of a data-set?</h3>

  • The mean of a data-set is given by the <u>sum of all values in the data-set, divided by the number of values</u>.
  • The standard deviation of a data-set is given by the <u>square root of the sum of the differences squared between each observation and the mean, divided by the number of values</u>.

When the measure of 84 is added to the data-set, we have that:

  • A measure greater than the mean is added, hence the mean would increase.
  • The difference squared between 84 and the mean of 81 is of 9, which is the same as the standard deviation, which would remain constant.

More can be learned about the mean and the standard deviation of a data-set at brainly.com/question/26941429

#SPJ1

8 0
1 year ago
[SCREENSHOT INCLUDED] If f(x) = sqrt(2x+1) then f '(4) =
Sphinxa [80]

Using the definition,

f'(4) = \displaystyle \lim_{x\to4} \frac{f(x)-f(4)}{x-4}

The numerator in the difference quotient is

f(x) - f(4) = \sqrt{2x+1} - \sqrt{2\cdot4+1} = \sqrt{2x+1}-3

Multiply the numerator and denominator by the conjugate of this expression,

\displaystyle \frac{\sqrt{2x+1}-3}{x-4} \cdot \dfrac{\sqrt{2x+1}+3}{\sqrt{2x+1}+3}

The modified numerator reduces to a difference of squares,

\displaystyle \frac{\left(\sqrt{2x+1}\right)^2-3^2}{(x-4)\left(\sqrt{2x+1}+3\right)}

Simplify this to get

\displaystyle \frac{2x+1-9}{(x-4)\left(\sqrt{2x+1}+3\right)} = \frac{2(x-4)}{(x-4)\left(\sqrt{2x+1}+3\right)}

Since x is approaching 4, it never actually takes on the value of 4, so (x - 4)/(x - 4) reduces to 1. Then the limit is equivalent to

f'(4) = \displaystyle \lim_{x\to4} \frac{f(x)-f(4)}{x-4} = \lim_{x\to4}\frac2{\sqrt{2x+1}+3}

and the remaining limand is continuous at x = 4, so that

f'(4) = \dfrac2{\sqrt{2\cdot4+1}+3} = \dfrac26 = \boxed{\dfrac13}

6 0
2 years ago
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