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3 years ago

arturo paid $8 in taxes on a purchase of $200. assuming the tac rate is the same, what would the tax be on a purchase of $150

Mathematics

1 answer:

Ludmilka [50]3 years ago

4 0

Answer:

you would take 1/4 away from 8 because 1/4 of $200 is 50 so that makes the taxes 2$

Step-by-step explanation:

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

HELP PLEASEEEEEEEEEEEEE

zhenek [66]

9514 1404 393

Answer:

2 1/12 days

Step-by-step explanation:

Time is the ratio of distance to speed.

(1200 mi)/(24 mi/h) × (1 day)/(24 h) = 25/12 day = 2 1/12 days

5 0

3 years ago

5) Solve. x - 32 = 73 A) x = 41 B) x = 105 C) x = 2336 D) x = 2 9/32

Mice21 [21]

Answer:

Step-by-step explanation:

105 - 32 = 73

6 0

3 years ago

Read 2 more answers

Michael jogs at 3 miles per hour and Jack runs at 4.5 miles per hour. If Michael starts at 2:00 pm and Jack starts on the same c

tresset_1 [31]

Answer:

It will take Jack 40 minutes to cover the same distance which Micheal covers in 60 minutes. Both will meet at 3 miles distance at 3 pm.

Step-by-step explanation:

Micheal runs 3 miles in one hour or 60 minutes.

In one minute he runs 3/60= 1/20 or 0.05miles.

Jack runs 4.5 miles in one hour or 60 minutes.

In one minute he runs 4.5/60= 45/600= 9/120= 3/40 or 0.075 miles.

So Jack is 0.075/0.05 = 1.5 times faster than Micheal.

or Micheal is 1.5 times slower than Jack.

Jack starts from 2:20 pm and at 3 pm he will run 0.075(40 minutes)= 3miles

Micheal start at 2:00 pm and at 3 pm he will run 0.05(60 minutes)= 3 miles

So at 3 pm both will have covered equal distance= 3miles.

It will take Jack 40 minutes to cover the same distance which Micheal covers in 60 minutes. Both will meet at 3 miles distance at 3 pm.

3 0

2 years ago

What is 35% of 60? just had to be 20 words

Vladimir [108]

Answer:

35% of 60 = 21

4 0

2 years ago

Read 2 more answers