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3 years ago

What are the first and last deciles of the standard Normal distribution?

Mathematics

1 answer:

aliya0001 [1]3 years ago

5 0

My best Guess is

1.282 for upper . , -1.282 for lower.

Hope this Helps

-Dante

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(8u+6v)(5u - Bv)?????

ivolga24 [154]

8ubv+6bv2+40u+30vu hope it’s right

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3 years ago

PLEASE SHOW WORK ILL GIVE BRAINLIEST DUE TODAY!!!

murzikaleks [220]

Answer:

1. 169 2. 163 3. 54 4. 221 5. 6 6. 7 7. 11 8. 9

Step-by-step explanation:

Remember the Order of Operations:

Parentheses

Exponents

Multiplication

Division

Addition

Subtract

*But always solve from left to right so there can be times where you either have to do division before multiplication or subtraction before addition

1. 14 + 18 ÷ 2 x 18 – 7

14+9 x 18-7

14+162-7

176-7

169

2. 15 x 10 + 12 ÷ 3 + 9

150+12÷3+9

150+4+9

154+9

163

3. 8 x 4 + 9 – 9 + 18

36+9-9+18

45-9+18

36+18

4. 2 - 1 + 5 x 4 x 11

2-1+20x11

2-1+220

1+220

221

5. 60 – 9 x 8 ÷ 8 x 6

60-72÷ 8 x 6

60-9x6

60-54

6. (10 ÷ 5)3 + 100 – 9 x 11

(2)3+100-9x11

6+100-9x11

6+100-99

106-99

7. 3 x 8 x 2 – 42 + 5

24x2-42+5

48-42+5

6+5

8. 14 ÷ 2 -1 + 3

7-1+3

6+3

6 0

3 years ago

You deposit $7200 in an account that pays 9.3% interest compounded yearly. Find the balance after 9 years.

inysia [295]

Answer:

Step-by-step explanation:

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2 years ago

A 12 m ladder leaning up against a wall makes a 50° angle with the

fiasKO [112]

Answer:

9.19m

Step-by-step explanation:

Check attachment

3 0

4 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago