All categories

3 years ago

What fraction of a dollar is represented by one dime

Mathematics

2 answers:

VLD [36.1K]3 years ago

8 0

10 over 100 is the answer

10/100

Simora [160]3 years ago

5 0

A dime is ten cents, and there are 100 cents in a dollar, so 10/100, or 1/10 when simplified. Hope it helps! If you could vote me brainiest, that would be awesome! Thanks!

You might be interested in

What is the equation for the line? Enter your answer in the box

Ipatiy [6.2K]

y=-1/2x-4

I could be wrong. Not my strong subject.

-Seth

4 0

4 years ago

Can someone help please Stickers are sold in packages of 15 sheets. If there are 20 students in the class, how many packages of

MrRa [10]

Answer:

4 packages

Step-by-step explanation:

If there are 20 and stickers are sold in 15 sheets, then you got to work backwards. take the answer and multiply it by the amount of sheets. For example: 4 times 15 is 60, 60 divided by 20 is 3. No leftover numbers.

8 0

3 years ago

Read 2 more answers

Chin-li is building a brick wall along the front of his property. The wall will have 15 rows of bricks, with 32 bricks in each r

Finger [1]

Answer:

<h3>The solution is 480 bricks.</h3>

Step-by-step explanation:

We are given that number of rows of bricks in a wall = 15 rows.

Number of bricks in a row = 32 bricks.

In order to find the total number of bricks, we need to multiply number of rows with number of bricks in a row.

Total number of bricks in 15 rows = 15 × 32 = 480 bricks.

Therefore, will Chin-li will need 480 bricks.

<h3>The solution is 480 bricks.</h3>

3 0

3 years ago

What is the equation of the circle with radius, r = 11 , and the center at (0, - 7)

Dimas [21]

Answer:

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given (h, k) = (0, - 7) and r = 11, then

(x - 0)² + (y - (- 7))² = 11², that is

x² + (y + 7)² = 121 → A

4 0

3 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

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

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$

80 miles:

This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

The probability of no damaging accidents is P(X = 0). So

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

6 0

3 years ago