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3 years ago

5

How do you find the tax on a purchase and calculate the final bill? Give an example, then write and solve a number sentence to i

llustrate your strategy.

1 answer:

Lerok [7]3 years ago

7 0

If you are buying something that cost $15 and sale tax is 6%, you can calculate final bill trough proportions:
$15:100%=$x:106%
100x=106*15, x=(106*15)/100, x=$15.9 is final bill.

You can also use easier way: $15+0.06*$15=$15.9
Explanation for 2nd way: Cost is $15, on that you need to add 6% of $15: (6%/100%)*$15 and that is 0.06*15.

You might be interested in

I'm pretty sure this is easy for some of you but not me!

saw5 [17]

Answer:

The possible outcomes are H1, H2, T1, and T2

Step-by-step explanation:

You can result in Heads and one, Heads and two, Tails and one, or Tails and two. Hope this helps!

5 0

3 years ago

Philip ran out of time while taking a multiple-choice test and plans to guess the last 4 questions. Each question has 5 possible

White raven [17]

Using the binomial distribution, it is found that there is a 0.4096 = 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

Considering that there are 4 questions, and each has 5 choices, the parameters are given as follows:

n = 4, p = 1/5 = 0.2.

The probability that he answers exactly 1 question correctly in the last 4 questions is P(X = 1), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{4,1}.(0.2)^{1}.(0.8)^{3} = 0.4096$

0.4096 = 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

8 0

2 years ago

Can you help me please?

9966 [12]

First, change the two mixed numbers in the expression into an improper fraction: 25/6 + 5/3.
Then, find the Lowest Common Denominator of both fractions, which is 6, and set both denominators equal to that. Remember, whatever you do on one side you must do to the other: 25/6 + 10/6
Add the two together: 25/6 +10/6 = 35/6.
To make it a mixed number again, find how many times 6 goes into 35, which is 5 times, with a remainder of 5. Your answer is 5 5/6

6 0

3 years ago

Melinda spend $14.25 on school supplies Dr spend $2.30 more than Melinda

zloy xaker [14]

14.25+2.30=16.55 so, dr spent $16.55 on school supplies

3 0

3 years ago

Read 2 more answers

I have an assignment and I am having trouble with it. Can someone please help ASAP???

bezimeni [28]

Answer:

A) Find the sketch in attachment.

In the sketch, we have plotted:

- The length of the arena on the x-axis (90 feet)

- The width of the arena on the y-axis (95 feet)

- The position of the robot at t = 2 sec (10,30) and its position at t = 8 sec (40,75)

The origin (0,0) is the southweast corner of the arena. The system of inequalities to descibe the region of the arena is:

$0\leq x \leq 90\\0\leq y \leq 95$

B)

Since the speed of the robot is constant, it covers equal distances (both in the x- and y- axis) in the same time.

Let's look at the x-axis: the robot has covered 10 ft in 2 s and 40 ft in 8 s. There is a direct proportionality between the two variables, x and t:

$\frac{10}{2}=\frac{40}{8}$

So, this means that at t = 0, the value of x is zero as well.

Also, we notice that the value of y increases by $\frac{75-30}{8-2}=7.5 ft/s$ (7.5 feet every second), so the initial value of y at t = 0 is:

$y(t=0)=30-7.5\cdot 2 =15 ft$

So, the initial position of the robot was (0,15) (15 feet above the southwest corner)

C)

The speed of the robot is given by

$v=\frac{d}{t}$

where d is the distance covered in the time interval t.

The distance covered is the one between the two points (10,30) and (40,75), so it is

$d=\sqrt{(40-10)^2+(75-30)^2}=54 ft$

While the time elapsed is

$t=8 sec-2 sec = 6 s$

Therefore the speed is

$v=\frac{54}{6}=9 ft/s$

D)

The equation for the line of the robot is:

$y=mx+q$

where m is the slope and q is the y-intercept.

The slope of the line is given by:

$m=\frac{75-30}{40-10}=1.5$

Which means that we can write an equation for the line as

$y=mx+q\\y=1.5x+q$

where q is the y-intercept. Substituting the point (10,30), we find the value of q:

$q=y-1.5x=30-1.5\cdot 10=15$

So, the equation of the line is

$y=1.5x+15$

E)

By prolonging the line above (40,75), we see that the line will hit the north wall. The point at which this happens is the intersection between the lines

$y=1.5x+15$

and the north wall, which has equation

$y=95$

By equating the two lines, we find:

$1.5x+15=95\\1.5x=80\\x=\frac{80}{15}=53.3 ft$

So the coordinates of impact are (53.3, 95).

F)

The distance covered between the time of impact and the initial moment is the distance between the two points, so:

$d=\sqrt{(53.5-0)^2+(95-15)^2}=95.7 ft$

From part B), we said that the y-coordinate of the robot increases by 15 feet/second.

We also know that the y-position at t = 0 is 15 feet.

This means that the y-position at time t is given by equation:

$y(t)=15+7.5t$

The time of impact is the time t for which

y = 95 ft

Substituting into the equation and solving for t, we find:

$95=15+7.5t\\7.5t=80\\t=10.7 s$

G)

The path followed by the robot is sketched in the second graph.

As the robot hits the north wall (at the point (53.3,95), as calculated previously), then it continues perpendicular to the wall, this means along a direction parallel to the y-axis until it hits the south wall.

As we can see from the sketch, the x-coordinate has not changed (53,3), while the y-coordinate is now zero: so, the robot hits the south wall at the point

(53.3, 0)

H)

The perimeter of the triangle is given by the sum of the length of the three sides.

- The length of 1st side was calculated in part F: $d_1 = 95.7 ft$

- The length of the 2nd side is equal to the width of the arena: $d_2=95 ft$

- The length of the 3rd side is the distance between the points (0,15) and (53.3,0):

$d_3=\sqrt{(0-53.3)^2+(15-0)^2}=55.4 ft$

So the perimeter is

$d=d_1+d_2+d_3=95.7+95+55.4=246.1 ft$

I)

The area of the triangle is given by:

$A=\frac{1}{2}bh$

where:

$b=53.5 ft$ is the base (the distance between the origin (0,0) and the point (53.3,0)

$h=95 ft$ is the height (the length of the 2nd side)

Therefore, the area is:

$A=\frac{1}{2}(53.5)(95)=2541.3 ft^2$

J)

The percentage of balls lying within the area of the triangle traced by the robot is proportional to the fraction of the area of the triangle with respect to the total area of the arena, so it is given by:

$p=\frac{A}{A'}\cdot 100$

where:

$A=2541.3 ft^2$ is the area of the triangle

$A'=90\cdot 95 =8550 ft^2$ is the total area of the arena

Therefore substituting, we find:

$p=\frac{2541.3}{8550}\cdot 100 =29.7\%$

4 0

3 years ago