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3 years ago

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X 3.6.PS-10 Question Help A new bank customer with $4,500 wants to open a money market account. The bank is offering a simple in

terest rate of 1.7%. a. How much interest will the customer earn in 10 years? b. What will the account balance be after 10 years?

1 answer:

makkiz [27]3 years ago

4 0

If it was in 10 years it would either be 2.6

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The temperature was -3° last night it is now -4° see what was the change in temperature was-3° C. It is -4° C. What was the chan

raketka [301]

Answer:

The temp got colder by -1

Step-by-step explanation:

The difference of -3 to -4

-4 + -3 = -1

7 0

3 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

2 years ago

Simplify 5x(3x^2+2x-3)

Tcecarenko [31]

Answer:

$15x^3 +10x^2 - 15x$

Step-by-step explanation:

Hello!

Use the distributive property and multiply like terms.

<h3>Simplify</h3>

$5x(3x^2 + 3x - 3)$
$5x(3x^2) + 5x(2x) + 5x(-3)$
$15x^3 +10x^2 - 15x$

The simplified form is $15x^3 +10x^2 - 15x$ .

4 0

1 year ago

Read 2 more answers

Please help!!!!!! I’ll mark Brainlyist!!!!!

Butoxors [25]

Answer:

its 55

Step-by-step explanation:

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2 years ago

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The boiling point of jet fuel is 329 degrees Fahrenheit. Rounded to the nearest degree, what is the temperature in degrees Celsi

djverab [1.8K]

Put the number in the formula and do the arithmetic.

C = (5/9)(329 - 32)

C = (5/9)(297)

C = 165

The temperature in °C is 165.

7 0

3 years ago