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3 years ago

11

An item that was selling for $72.00 is reduced to 60.00. Find the percent decrease in price. Round your answer to the nearest te

nth

1 answer:

Anestetic [448]3 years ago

4 0

The answer is 16.7 percent

You might be interested in

Which animal traveled at a rate of 1.5 meters per minute?

Fofino [41]

Answer:

The ant

Step-by-step explanation:

We know that the meters traveled is going to be more than the amount of minutes because it is 1.5:1 so that narrows it down to the turtle and the ant.

Now we can just take 150% of each and see if they match up.

150% of 5 is 7.5 so therefore it is not the Turtle which means it must be the ant.

Hope this helped ! !

<3

if it did please consider marking brainliest i would appreciate it greatly :)

5 0

2 years ago

Read 2 more answers

9. Find the length of a base of a triangle with<br>an Area of 100 and a height of 25.

liraira [26]

Answer:

8 (units)

Step-by-step explanation:

The formula to find the area of a triangle is A = 1/2(b*h). Since we know the area already, we can substitute that value in to calculate the answer backwards.

Plug in the numbers known:

A = 100

h = 25

Equation: 100 = 1/2(b*25)

Now solve:

100 = 1/2(b*25)

200 = b * 25

<em>8 = b</em>

So, the answer to your question is 8 (units).

Hope this helps!

6 0

3 years ago

Sergio is working on a research project involving the ages of students at Brooklyn College. After interviewing 500 students, he

Zinaida [17]

Answer:

The probability of of a randomly chosen student being exactly 21 years old.

= 1.293

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given Population size n = 500</em>

<em>Mean of the Population = 20 years and 6 months</em>

<em> = </em> $20 + \frac{6}{12} = 20 +0.5 = 20.5 years$ <em></em>

<em>Standard deviation of the Population = 2 years</em>

Let 'X' be the range of ages of the students on campus follows a normal distribution

Let x =21

$Z = \frac{x-mean}{S.D}$

$Z = \frac{x-mean}{S.D} = \frac{21-20.6}{2} =0.2$

<em>The probability of a randomly chosen student being exactly 21 years old.</em>

<em>P( Z≤21) = 0.5 + A( 0.2) </em>

= 0.5 +0.793

= 1.293

6 0

3 years ago

Most college-bound students take either the SAT(Scholastic Assessment Test) or the ACT (which originally stood for American coll

Jlenok [28]

Answer:

Luis would need to have a SAT score of 574.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Nicole's z-score:

ACT scores have a mean of about 21 with a standard deviation of about 5, which means that $\mu = 21, \sigma = 5$

Nicole gets a score of 24, which means that $X = 24$ . Her z-score is:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24 - 21}{5}$

$Z = 0.6$

What score would Luis have to have on the SAT to have the same standardized score(z-score) as Nicole's standardized score on the ACT?

Luis would have to get a score with a z-score of 0.6, that is, X when Z = 0.6.

SAT scores have a mean of about 508 with a standard deviation of about 110, which means that $\mu = 508, \sigma = 110$ .

The score is:

$Z = \frac{X - \mu}{\sigma}$

$0.6 = \frac{X - 508}{110}$

$X - 508 = 0.6*110$

$X = 574$

Luis would need to have a SAT score of 574.

8 0

2 years ago

A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of

ddd [48]

There are

$\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}$

ways of pairing up any 2 members from the pool of $2n$ contestants. Note that

$(2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)$

so that

$\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}$

4 0

1 year ago