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2 years ago

6

How many dimes and nickels make 6.10 dollars if there are 79 coins

2 answers:

Tamiku [17]2 years ago

8 0

But in all 67 dimes and nickels put together

7nadin3 [17]2 years ago

4 0

Answer:

50 dimes and 29 nickels

Step-by-step explanation:

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(2^4x+4)÷(2^x+1)<br> 60 POINTS<br> BRANLIEST IF RIGHT

zmey [24]

Answer:

$\frac{2^4^x + 4}{2^x + 1}$

The file includes the input of your equation, comment if the equation you input is a mistake.

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2 years ago

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What is the midpoint of the segment shown below?

Aneli [31]

Answer:

(1, -3/2)

Step-by-step explanation:

The x coordinate is the same for both endpoints so the x coordinate for the midpoint is 1

The y coordinate for the midpoint is found by adding the two y coordinates and dividing by 2

(2+-5)/2 = -3/2

The midpoint is

(1, -3/2)

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3 years ago

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3)Find the GCF and LCM of 10 and 28. Show your work.<br><br> HELP LAST QUESTION!!!!!

soldier1979 [14.2K]

Answer:

GCF=2 LCM=140

Step-by-step explanation:

8 0

2 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

<img src="https://tex.z-dn.net/?f=5%20%2B%20%20%5Cfrac%7Bx%20-%202%7D%7B3%7D%20%20%3D%20%20%5Cfrac%7Bx%20%2B%203%7D%7B8%7D%20" i

sammy [17]

First multiply each term by 24(the common denominator found by 8×3) to remove the fractions and make things easier.
$24(5) + 24( \frac{x - 2}{3} ) = 24( \frac{x + 3}{8} )$
This will give you,
$120 + 8(x - 2) = 3(x + 3)$
Then just continue to simplify and isolate the variable.
$120 + 8x - 16 = 3x + 9$
$8x + 104 = 3x + 9$
$5x = - 95$
$x = - 19$

7 0

2 years ago