3 years ago

A student’s standardised mark on a class test was z = –1.7. The mean mark for the class is 55 and the standard

1 answer:

7 0

The answer is 58, standard deviation is basically telling you that the space between each point in your statistics is on average that amount so that can tell us to take the mean mark of 55 and add 2.85 = 57.85

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Let f(y) = y^4 - 3y^3 + y - 3 and g(y) = y^3 + 7y^2 - 2. Find f(y) + g(y). Write your answer as a polynomial with terms of decre

Anarel [89]

<h3>Answer: y^4 - 2y^3 + 7y^2 + y - 5</h3>

Work Shown:

h(y) = f(y) + g(y)

h(y) = (y^4 - 3y^3 + y - 3) + (y^3 + 7y^2 - 2)

h(y) = y^4 - 3y^3 + y - 3 + y^3 + 7y^2 - 2

h(y) = y^4 + (-3y^3+y^3) + 7y^2 + y + (-3 - 2)

h(y) = y^4 - 2y^3 + 7y^2 + y - 5

This is a fourth degree polynomial (aka quartic).

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3 years ago

Which of the following pieces are congruent?

posledela

Answer:

The horizontal segment of E and vertical segments of M.

Step-by-step explanation:

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3 years ago

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80,607.202 rounded to the nearest tens place is

postnew [5]

It’s 80,610 if you really mean the tens place.
If you mean the one tenths place then it’s 80,607.2

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3 years ago

Find two positive numbers whose product is 25 and whose sum is a minimum. (if both values are the same number, enter it into bot

pshichka [43]

Let xy = 25, x and y > 0.

$y = \frac{25}{x}$
$\text{Denote S as: } x + y$
$S(x) = x + \frac{25}{x}$

$\text{Minimum: }S'(x) = 0, S''(x) > 0$
$S'(x) = 1 - \frac{25}{x^{2}}$
$1 - \frac{25}{x^{2}} = 0$
$x^{2} = 25$
$x = 5 (x > 0)$

$S''(x) = \frac{50}{x^{3}} > 0, (x > 0)$

Thus, at x = 5, there lies a minimum summation, and this forms a square with side length of 5.

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4 years ago

1/2 of 4/5 as a fraction<br><br>

omeli [17]

Answer:

4/10 is the answer

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2 years ago

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