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Mathematics College

There are 30 students in Mrs. Woodward’s class, and 1/5 of the class has their own cell phone. Of this group of students, 1/2 of them are allowed to use social media. How many of the students have a cell phone and can use social media?

2 months ago

Solution 1

Guest #1627964

2 months ago

Answer:

3 students

Step-by-step explanation:

There are 30 students in Mrs. Woodward’s class.

$\dfrac{1}{5}$ of the class has their own cell phone, so

$\dfrac{1}{5}\cdot 30=\dfrac{1}{5}\cdot \dfrac{30}{1}=6$

students have their own cell phones.

$\dfrac{1}{2}$ of those 6 students are allowed to use social media. So,

$\dfrac{1}{2}\cdot 6=\dfrac{1}{2}\cdot \dfrac{6}{1}=3$

students are allowed to use social media.

📚 Related Questions

Question

The College Boards, which are administered each year to many thousands of high school students, are scored so as to yield a mean of 513 and a standard deviation of 130. These scores are close to being normally distributed. What percentage of the scores can be expected to satisfy each of the following conditions? a. Greater than 600 b. Greater than 700 c. Less than 450 d. Between 450 and 600 Bus. 4.71 Monthly sales figures for a particular

Solution 1

Answer:

a) 25.14% percentage of the scores expected to be greater than 600.

b) 7.49% percentage of the scores expected to be greater than 700.

c) 31.56% of the scores are expected to be less than 450.

d) 43.3% of the scores are expected to be between 450 and 600.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

For this problem, we have that:

The College Boards, which are administered each year to many thousands of high school students, are scored so as to yield a mean of 513 and a standard deviation of 130, so $\mu = 513, \sigma = 130$ .

What percentage of the scores can be expected to satisfy each of the following conditions?

a) Greater than 600

This is 1 subtracted by the pvalue of Z when $X = 600$

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

$Z = \frac{600 - 513}{130}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

So there is a 1 - 0.7486 = 0.2514 = 25.14% percentage of the scores expected to be greater than 600.

b) Greater than 700

This is 1 subtracted by the pvalue of Z when $X = 700$

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

$Z = \frac{700 - 513}{130}$

$Z = 1.44$

$Z = 1.44$ has a pvalue of 0.9251.

So there is a 1 - 0.9251 = 0.0749 = 7.49% percentage of the scores expected to be greater than 700.

c. Less than 450

This is the pvalue of Z when $X = 450$

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

$Z = \frac{450 - 513}{130}$

$Z = -0.48$

$Z = -0.48$ has a pvalue 0.31561.

So, 31.56% of the scores are expected to be less than 450.

d. Between 450 and 600

This is the subtraction of the pvalue of $X = 600$ by the pvalue of $X = 450$

$P = 0.7486 - 0.31561 = 0.433$

43.3% of the scores are expected to be between 450 and 600.

Question

70%70, percent of a number is what fraction of that number?

Solution 1

Answer:

7/10

Step-by-step explanation:

In order to solve this you just have to make the division of 70 by 100, and thenn reduce that as much as you possibly can:

$\frac{70}{100} \\\frac{35}{50}\\\frac{7}{10}$

So now you know that the fraction of a number that 70% of that number represents is 7/10.

Solution 2

Hello there!

A number that is 70% of a number is 7/10 of that number in fraction form!

I hope this was helpful and have a great rest of your day!

Question

On a coordinate plane, triangle A B C is shown. Point A is at (negative 1, 3), point B is at (negative 5, negative 1), and point C is at (3, negative 1). What is true about △ABC? Select three options AB ⊥ AC The triangle is a right triangle. The triangle is an isosceles triangle. The triangle is an equilateral triangle. BC ∥ AC

Solution 1

Answer:

Option C, The triangle is an isosceles triangle.

Step-by-step explanation:

On a coordinate plane, triangle A B C is shown. Point A is at (negative 1, 3), point B is at (negative 5, negative 1), and point C is at (3, negative 1).

$\left(-1,3\right), \left(-5,-1\right),\left(3,-1\right)$

Use distance formula to find $AB^2 , BC^2, CA^2\\AB^2 = (-1+5)^2+(3+1)^2 =32\\AC^2= ((-1-3)^2+(3+1)^2 = 32\\BC^2 = (-5-3)^2+(-1+1)^2 = 64$

We find only two sides are equal and third side is not equal

Hence this is an isosceles triangle

Solution 2

Answer:

Options A, B, and C

Step-by-step explanation:

Power of Taking the Test

Question

One ratio you could use for converting units is 1 kilograr 1,000 grarns O A. True OB. False

Solution 1

Answer:

true

Step-by-step explanation:

The international system of measures established for units that have to do with weight the standard measure would be in grams, to express larger units to avoid very long numbers it was established that one kilogram equals 1000g, so the ratio between them is 1000 .

to convert from grams to kilograms we divide by 1000.

to convert kilograms to grams multiply by 1000

Solution 2

A. True

................

Question

Pat Smith and Kelly Jones are engaged. What are possible last name combinations for the married couple (listing Pat first)

Solution 1

Answer:

just one

Step-by-step explanation:

if it is established that the last names of pat smith go first than those of kelly jones and each one has only one lastname as indicated by the problem, there is only one possibility that the last name for the married couple is the combination "smith jones"

Question

Simplify the fourth root of three over the fifth root of three.

Solution 1

In simplified form the expression can be written as $\rm 3^{\frac{1}{20}$ .

What is an Expression ?

An expression is a mathematical statement consisting of variables , coefficients and mathematical operators.

An expression is given

$\rm \dfrac{\sqrt[4]{3} }{\sqrt[5]{3}}$

Converting them into exponent form , the expression can be written as

$\rm \dfrac{ 3 ^{1/4}}{3^{1/5}}\\\\a^m / a^n = a ^{m-n}$

Therefore simplifying

$\rm 3^{\frac{1}{4} - \frac{1}{5} \\\\\\\\3^{\frac{1}{20}$

Therefore In simplified form the expression can be written as $\rm 3^{\frac{1}{20}$ .

To know more about Expression

brainly.com/question/14083225

#SPJ2

Solution 2

Answer:

$\sqrt[20]{3}$

Step-by-step explanation:

$\sqrt[4]{3}$ / $\sqrt[5]{3}$

= $3^{1/4}$ / $3^{1/5}$

= $3^{(1/4) - (1/5)}$

= $3^{(1/20)}$ (= $\sqrt[20]{3}$ )

Question

5 decreased by 4 times m

Solution 1

Answer:

5 - 4m

Step-by-step explanation:

Question

The quotient of x and 2 increased by 7

Solution 1

Answer:

$x \div 2 + 7$

Is this what you were looking for?

Solution 2

Answer:

x/2 + 7

Step-by-step explanation:

First you need to figure out the equation.

Quotient means you are dividing something, so you are dividing x and 2.

Increased means you are adding something, so somewhere in the equation you will add 7.

Here is what the equation would be:

x/2 + 7

Question

Pat and Sam are arguing about the probability of obtaining 2 heads and 2 tails in 4 flips of a fair coin, in that specific order. Pat says, "The probability equals 1/16, since there is one way this outcome can occur, and the experiment has 16 different possible outcomes". Sam says "The probability equals 6/16, since the corresponding entry in row 4 of Pascal's triangle is 6, and the sum of the entries in that row equals 16." Who is correct, and why?

Solution 1

Answer:

Pat's right, because there are no permutations.

Step-by-step explanation:

Since there is a specific order, there are no permutations. The relationship between the binomial distribution and Pascal's triangle happens when there are permutations.

So, the correct logic is Pat's.

Each flip has 0.5 probability of having the desired result. There are four flips. So:

$P = (0.5)^{4} = \frac{1}{2}^{4} = \frac{1}{16}$ .

Question

Suppose that 681 tennis players want to play an elimination tournament. That means: they pair up, at random, for each round; if the number of players before the round begins is odd, one of them, chosen at random, sits out that round. The winners of each round, and the odd one who sat it out (if there was an odd one), play in the next round, till, finally, there is only one winner, the champion. What is the total number of matches to be played altogether, in all the rounds of the tournament?

Solution 1

Answer:

680 matches are played altogether.

Step-by-step explanation:

A number is odd if the rest of the division of the number by 2 is one.

So, for each round, we have that the number of matches played is the number of players in the start of the round divided by 2.

The number of players at the end of the round is the number of matches(each match has a winner, that remains in the tournament) plus the rest of the division(the odd player that sit out the round).

First round:

681 players

681/2 = 340 mod 1

Number of matches in round: 340

Total number of matches: 340

Number of players at the end of the round: 340 + 1 = 341

Second round

341 players

341/2 = 170 mod 1

Number of matches in round: 170

Total number of matches: 340 + 170 = 510

Number of players at the end of the round: 170 + 1 = 171

Third round

171 players

171/2 = 85 mod 1

Number of matches in round: 85

Total number of matches: 510 + 85 = 595

Number of players at the end of the round: 85 + 1 = 86

Fourth round

86 players

86/2 = 43 mod 0

Number of matches in round: 43

Total number of matches: 595 + 43 = 638

Number of players at the end of the round: 43 + 0 = 43

Fifth round

43 players

43/2 = 21 mod 1

Number of matches in round: 21

Total number of matches: 638 + 21 = 659

Number of players at the end of the round: 21 + 1 = 22

Sixth round

22 players

22/2 = 11 mod 0

Number of matches in round: 11

Total number of matches: 659+11 = 670

Number of players at the end of the round: 11 + 0 = 11

Seventh round

11 players

11/2 = 5 mod 1

Number of matches in round: 5

Total number of matches: 670+5 = 675

Number of players at the end of the round: 5 + 1 = 6

Eight round

6 players

6/2 = 3 mod 0

Number of matches in round: 3

Total number of matches: 675+3 = 678

Number of players at the end of the round: 3 + 0 = 3

Ninth round

3 players

3/2 = 1 mod 1

Number of matches in round: 1

Total number of matches: 678+1 = 679

Number of players at the end of the round: 1 + 1 = 2

Tenth round

2 players

2/2 = 1 mod 0

Number of matches in round: 1

Total number of matches: 679+1 = 680

Number of players at the end of the round: 1 + 0 = 1(the champion)

So, 680 matches are played altogether.

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