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2 years ago

The tax rate of $.6943 in decimal form can be expressed per $100 as A. $6.943. B. $690.3. C. $69.43. D. $69.43 mills.

Mathematics

2 answers:

hichkok12 [17]2 years ago

5 0

Answer:

The answer is option D. $69.43

Step-by-step explanation:

Given tax rate per $100 = $0.6943

So, we will multiply this rate with 100.

$.6943\times100$ = $69.43

The tax rate here, is the per dollar times per 100 of the assessed value.

Therefore, the answer is option D. $69.43

erica [24]2 years ago

4 0

Let
x-------> <span>The tax rate in decimal form

we know that
x*100=6943
divide by 100 both sides
x=6943/100
x=69.43

the answer is the option
</span><span>C. $69.43</span>

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The local orchestra has been invited to play at a festival. There are 111 members of the orchestra and 6 are licensed to drive l

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- Make an equation representing the number of vehicles needed.

We have six drivers so

x + y ≤ 6

That's not really an equation; it's an inequality. We want to use all our drivers so we can use the small vans, so

x + y = 6

- Make an equation representing the total number of seats in vehicles for the orchestra members.

s = 25x + 12y

That's how many seats total; it has to be at least 111 so again an inequality,

25x + 12y ≥ 111

We solve it like a system of equations.

x + y = 6

y = 6 - x

111 = 25x + 12y = 25x + 12(6-x)

111 = 25x + 72 - 12x

111 - 72 = 13 x

39 = 13 x

x = 3

Look at that, it worked out exactly. It didn't have to.

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Answer: 3 buses, 3 vans

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2 years ago

Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum

dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

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2 years ago

A bucket that has a mass of 30 kg when filled with sand needs to be lifted to the top of a 30 meter tall building. You have a ro

Rom4ik [11]

Answer:

765 J

Step-by-step explanation:

We are given;

Mass of bucket = 30 kg

Mass of rope = 0.3 kg/m

height of building= 30 meter

Now,

work done lifting the bucket (sand and rope) to the building = work done in lifting the rope + work done in lifting the sand

Or W = W1 + W2

Work done in lifting the rope is given as,

W1 = Force x displacement

W1 = (30,0)∫(0.2x .dx)

Integrating, we have;

W1 = [0.2x²/2] at boundary of 30 and 0

W1 = 0.1(30²)

W1 = 90 J

work done in lifting the sand is given as;

W2 = (30,0)∫(F .dx)

F = mx + c

Where, c = 30 - 15 = 15

m = (30 - 15)/(30 - 0)

m = 15/30 = 0.5

So,

F = 0.5x + 15

Thus,

W2 = (30,0)∫(0.5x + 15 .dx)

Integrating, we have;

W2 = (0.5x²/2) + 15x at boundary of 30 and 0

So,

W2 = (0.5 × 30²)/2) + 15(30)

W2 = 225 + 450

W2 = 675 J

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,

W = 90 + 675

W = 765 J

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2 years ago

Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

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