answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
yuradex [85]
2 years ago
9

Zeke goes to a skate shop with his friend Tristan. Tristan buys a helmet that costs $28.75. Zeke finds a helmet that is

="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B5%7D" id="TexFormula1" title="\frac{3}{5}" alt="\frac{3}{5}" align="absmiddle" class="latex-formula"> of the cost of Tristan's helmet. If Zeke gives the cashier a $20 bill, how much change will he receive.
Mathematics
1 answer:
Lemur [1.5K]2 years ago
4 0

Answer: $2.75

Step-by-step explanation:

Zeke's helmet cost 3/5 x 28.75 = 17.25

Change = 20 - 17.25 = 2.75

You might be interested in
The graph shows the increase of cost of strawberries per pound in dollars, y, over a period of years, x. What is the rate of inc
wel

Answer:

the answer is $0.50

6 0
1 year ago
Read 2 more answers
The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Nataliya [291]

Answer:

(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716

(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;

    Mean, \mu = 30.05 inches        and    Standard deviation, \sigma = 0.2 inches

Let X = A sheet selected at random from the population

Here, the standard normal formula is ;

                  Z = \frac{X - \mu}{\sigma} ~ N(0,1)

(a) <em>The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65) </em>

P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)

P(X < 30.65) = P(\frac{X - \mu}{\sigma} < \frac{30.65 - 30.05}{0.2} ) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425

                                                                                                = 0.9985

P(X <= 30.25) = P( \frac{X - \mu}{\sigma} <= \frac{30.25 - 30.05}{0.2} ) = P(Z <= 1) = 0.84134

Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .

(b)<em> Let Y = Standard Normal Variable is given by N(0,1) </em>

<em> Which means mean of Y = 0 and standard deviation of Y = 1</em>

So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)

 P(Y < 3.2) = P(\frac{Y - \mu}{\sigma} < \frac{3.2 - 0}{1} ) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688

                                                                                           = 0.99931

 P(Y <= 0.3) = P(\frac{Y - \mu}{\sigma} <= \frac{0.3 - 0}{1} ) = P(Z <= 0.3) = 0.61791

Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .

 

3 0
2 years ago
A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s
amid [387]

Answer:

The correct conclusion is:

<em>"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."</em>

Step-by-step explanation:

A doctor is measuring the average height of male students at a large college.

The doctor measures the heights, in inches, of a sample of 40 male students from the baseball team.

Using this data, the doctor calculates the 95% confidence interval (63.5, 74.4).

The following conclusions is valid:

<em>"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."</em>

Since we know that the confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.  

For the given case, the confidence level is 95% and the corresponding confidence interval is (63.5, 74.4) which represents the true mean of heights for male students at the college where the doctor measured heights.

Therefore, it is valid to conclude that the doctor is 95% confident that the mean height of male students at the college is within the interval of (63.5, 74.4).

6 0
2 years ago
33s^5t^4u^8 ÷ 11st^3u^8
miskamm [114]

Answer:

3s^{4}t\\

Step-by-step explanation:

7 0
1 year ago
The recipe has 10 tomatoes, 6 cucumbers, and 3 peppers. What is the ratio of tomatoes to peppers?
Tema [17]

Answer:

10 : 3

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Other questions:
  • If the m&lt;7 = 115 degrees, find the measure of &lt;2
    10·2 answers
  • In a batch of 100 cell phones, there are, on average, 5 defective ones. if a random sample of 30 is selected, find the probabili
    5·1 answer
  • PLEASE HELP NOW 20pts........
    9·2 answers
  • If the x- and y-values in each pair of a set of ordered pairs are interchanged, the resulting set of ordered pairs is known as t
    6·1 answer
  • If a hurricane was headed your way, would you evacuate? The headline of a press release states, "Thirty-four Percent of People o
    6·1 answer
  • When a certain basketball player takes his first shot in a game he succeeds with probability 1/2. If he misses his first shot, h
    12·1 answer
  • A group of high school seniors took a scholastic aptitude test. The resulting math scores had a mean 504.7 with a standard devia
    11·1 answer
  • Which of the following appear in the diagram below? Check ALL that apply.
    11·2 answers
  • 1. The speed at which an automated assembly line produces a product follows a normal distribution with mean production time of 3
    5·1 answer
  • 11. In a certain town the ratio of the number of cars to that of taxis is 15:4. If
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!