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3 years ago

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Suppose that IQ scores have a bell-shaped distribution with a mean of 104 and a standard deviation of 17. Using the empirical ru

le, what percentage of IQ scores are between 87 and 121

1 answer:

Shalnov [3]3 years ago

4 0

Answer:

By the Empirical Rule, 68% of IQ scores are between 87 and 121

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 104

Standard deviation = 17

Using the empirical rule, what percentage of IQ scores are between 87 and 121

87 = 104 - 1*17

So 87 is one standard deviation below the mean

121 = 104 + 1*17

So 121 is one standard deviation above the mean

By the Empirical Rule, 68% of IQ scores are between 87 and 121

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One angle measures 27° more than 2 times another. If the two angles are complementary, find the measures of the angles.

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Answer:

<em>A. 21°, 69°</em>

Step-by-step explanation:

If you work by process of elimination all you have to do is take 27 away from the bigger degree of the two and see if it is 2x as much as the smaller degree.

Ex.

1. 69°-27°= 42°, which is 2x as many as 21°.

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How to write 3,000,000+600,000+80,000+10 in written form

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Written form is just how you would usually see a number. So it would be 3,680,010.

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3 years ago

Read 2 more answers

Solve each inequality, and then drag the correct solution graph to the inequality.

Nesterboy [21]

The correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

(NOTE: The graphs are labelled A, B and C from left to right)

For the first inequality,

$4(9x-18)>3(8x+12)$

First, clear the brackets,

$36x-72>24x+36$

Then, collect like terms

$36x-24x>36+72\\12x >108$

Now divide both sides by 12

$\frac{12x}{12} > \frac{108}{12}$

∴ $x > 9$

For the second inequality

$-\frac{1}{3}(12x+6) \geq -2x +14$

First, clear the fraction by multiplying both sides by 3

$3 \times[-\frac{1}{3}(12x+6)] \geq3 \times( -2x +14)$

$-1(12x+6) \geq -6x +42$

Now, open the bracket

$-12x-6 \geq -6x +42$

Collect like terms

$-6 -42\geq -6x +12x$

$-48\geq 6x$

Divide both sides by 6

$\frac{-48}{6} \geq \frac{6x}{6}$

$-8\geq x$

∴ $x\leq -8$

For the third inequality,

$1.6(x+8)\geq 38.4$

First, clear the brackets

$1.6x + 12.8\geq 38.4$

Collect likes terms

$1.6x \geq 38.4-12.8$

$1.6x \geq 25.6$

Divide both sides by 1.6

$\frac{1.6x}{1.6}\geq \frac{25.6}{1.6}$

∴ $x \geq 16$

Let the graphs be A, B and C from left to right

The first graph (A) shows $x\leq -8$ and this matches the 2nd inequality

The second graph (B) shows $x \geq 16$ and this matches the 3rd inequality

The third graph (C) shows $x > 9$ and this matches the 1st inequality

Hence, the correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

Learn more here: brainly.com/question/17448505

8 0

2 years ago

I'm confused. Can someone help?

VARVARA [1.3K]

a.

The polynomial w^2+18w+84 cannot be factored

The perfect square trinomial is w^2+18w + 81

----------

The reason the original can't be factored is that solving w^2+18w+84=0 leads to no real solutions. Use the quadratic formula to see this. The graph of y = x^2+18x+84 shows there are no x intercepts. A solution and an x intercept are basically the same. The x intercept visually represents the solution.

w^2+18w+81 factors to (w+9)^2 which is the same as (w+9)(w+9). We can note that w^2+18w+81 is in the form a^2+2ab+b^2 with a = w and b = 9

================================================

b.

The polynomial y^2-10y+23 cannot be factored

The perfect square trinomial is y^2-10y + 25

---------

Using the quadratic formula, y^2-10y+23 = 0 has no rational solutions. The two irrational solutions mean that we can't factor over the rationals. Put another way, there are no two whole numbers such that they multiply to 23 and add to -10 at the same time.

If we want to complete the square for y^2-10y, we take half of the -10 to get -5, then square this to get 25. Therefore, y^2-10y+25 is a perfect square and it factors to (y-5)^2 or (y-5)(y-5)

8 0

3 years ago