A 500.0 g piece of aluminum at 100° C is placed in 300ml of water. While in the water, the

aluminum then cools to 30°C. Calculate the amount of heat lost by the aluminum. The

specific heat of water is 4.18 J/g °C and the specific heat of aluminum is 0.90 J/g °C

Answers

Answer 1

Answer:

The amount of heat lost by the aluminum is 31,500 J

Step-by-step explanation:

Given;

mass of aluminum, m =  500 g

initial temperature of the aluminum, θ₁ = 100° C

final temperature of the aluminum, θ₂ = 30°C

specific heat capacity of water, C = 4.18 J/g °C

specific heat capacity of aluminum , C = 0.90 J/g

Heat lost by the aluminum is equal to heat gained by the water.

The amount of heat lost by the aluminum, is calculated as;

Q = MCΔθ

Q = 500 x 0.9 (100 - 30)

Q = 500 x 0.9 x 70

Q = 31,500 J

Therefore, the amount of heat lost by the aluminum is 31,500 J


Related Questions

Find an Equation of a line with the given slope that passes through the point. Write the equation in the form Ax + By=C
M=3/2, (7,-2) -problem
Bridge math sails
Module 4B2

Answers

Answer:

c = 24 can i get brainliest

Step-by-step explanation:

Find the general solution to y′′+6y′+13y=0. Give your answer as y=.... In your answer, use c1 and c2 to denote arbitrary constants and x the independent variable. Enter c1 as c1 and c2 as c2.

Answers

Answer:

[tex]y(x)=c_1e^{-3x} cos(2x)+c_2e^{-3x} sin(2x)[/tex]

Step-by-step explanation:

In order to find the general solution of a homogeneous second order differential equation, we need to solve the characteristic equation. This is basically as easy as solving a quadratic.

For a second order differential equation of type:

[tex]ay''+by'+cy=0[/tex]

Has characteristic equation:

[tex]a r^{2} +br+c=0[/tex]

Whose solutions [tex]r_1 , r_2 ,.., r_n[/tex] are the roots from which the general solution can be formed. There are three cases:

Real roots:

[tex]y(x)=c_1e^{r_1 x} +c_2e^{r_2 x}[/tex]

Repeated roots:

[tex]y(x)=c_1e^{r x} +c_1 xe^{r x}[/tex]

Complex roots:

[tex]y(x)=c_1e^{\lambda x}cos(\mu x) +c_2e^{\lambda x}sin(\mu x)\\\\Where:\\\\r_1_,_2=\lambda \pm \mu i[/tex]

Therefore:

The characteristic equation for:

[tex]y''+6y'+13y=0[/tex]

Is:

[tex]r^{2} +6r+13=0[/tex]

Solving for [tex]r[/tex] :

[tex]r_1_,_2= -3 \pm 2i[/tex]

So:

[tex]\mu = 2\\\\and\\\\\lambda=-3[/tex]

Hence, the general solution of the differential equation will be given by:

[tex]y(x)=c_1e^{-3x} cos(2x)+c_2e^{-3x} sin(2x)[/tex]

Please answer this correctly

Answers

Answer:

Bailey: 16%

Coco: 28%

Ginger: 32%

Ruby: 24%

I hope this helps!

Find the constant of variation for the relation and use it to write and solve the equation.

if y varies directly as x and as the square of z, and y=25/9 when x=5 and z=1, find y when x=1 and z=4

Answers

Answer:

When x = 1 and z = 4,   [tex]y=\frac{80}{9}[/tex]

Step-by-step explanation:

The variation described in the problem can be written using a constant of proportionality "b" as:

[tex]y=b\,\,x\,\,z^2[/tex]

The other piece of information is that when x = 5 and z = 1, then y gives 25/9. So we use this info to find the constant "b":

[tex]y=b\,\,x\,\,z^2\\\frac{25}{9} =b\,\,(5)\,\,(1)^2\\\frac{25}{9} =b\,\,(5)\\b=\frac{5}{9}[/tex]

Knowing this constant, we can find the value of y when x=1 and z=4 as:

[tex]y=b\,\,x\,\,z^2\\y=\frac{5}{9} \,\,x\,\,z^2\\y=\frac{5}{9} \,\,(1)\,\,(4)^2\\y=\frac{5*16}{9}\\y=\frac{80}{9}[/tex]

On Sunday, a local hamburger shop sold a combined of 572 hamburger and cheeseburger. The number of cheeseburgers sold was three times the number of hamburger sold. How many hamburger were sold on Sunday

Answers

Answer: 143 hamburgers and 429 cheese burgers

Explanation:

Call h and c the number of both items.
(h-hamburger and c-cheeseburger)

h + c = 572
c = 3h

Sub the second into the first

h + 3h = 572
4h = 572

Divide both sides by 4
h = 143 hamburgers

Use this back into the second equation
c = 3 • 143 = 429 cheeseburgers

If you have changed the tires on your car, the original diameter is 24.5 inches. to a new diameter of 26 inches, how fast are you actually going if your speedometer is reading 53 mph? A. 50.5 mph B. 53 mph C. 56.2 mph D. 62.8 mph

Answers

Answer:  c) 56.2

Step-by-step explanation:

Compare the original rate to the the new rate:

[tex]\dfrac{diameter}{mph}:\dfrac{24.5\ in}{53\ mph}=\dfrac{26\ in}{x}\\\\\\24.5x=53(26)\\\\\\x=\dfrac{53(26)}{24.5}\\\\\\x=\large\boxed{56.2\ mph}[/tex]

The following histogram shows the exam scores for a Prealgebra class. Use this histogram to answer the questions.Prealgebra Exam ScoresScores 70.5, 75.5, 80.5, 85.5, 90.5, 95.5, 100.5Frequency 0, 4, 8, 12, 16, 20, 24Step 1 of 5:Find the number of the class containing the largest number of exam scores (1, 2, 3, 4, 5, or 6).Step 2 of 5:Find the upper class limit of the third class.Step 3 of 5:Find the class width for this histogram.Step 4 of 5:Find the number of students that took this exam.Step 5 of 5:Find the percentage of students that scored higher than 95.595.5. Round your answer to the nearest percent.

Answers

Answer:

The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

For the third class 78 - 83 ; the upper limit = 83

The class width for this histogram 5

The number of students that took the exam simply refers to the frequency is  84

The percentage of students that scored higher than 95.5 is 53%

Step-by-step explanation:

The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:

NOW;

The table given in the question can be illustrated as follows:

S/N           Class                Score          Frequency

1                 68  - 73            70.5           0

2                73 - 78             75.5           4

3                78 - 83             80.5           8

4                83 - 88             85.5           12

5                88 - 93            90.5           16

6                93 - 98            95.5           20

7                98 - 103           100.5          24                            

TOTAL:                                                 84

                                                                                           

a) The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

b) For the third class 78 - 83 ; the upper limit = 83  ( since the upper limit is derived by addition of  5 to the last number showing  in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))

c)   The class width for this histogram 5 ; since it  is the difference  between the upper and lower boundaries  limit of the given class.

So , from above the difference in any of the class will definitely result into 5

d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84

e) Lastly; the percentage of students that scored higher than 95.5 is ;

⇒[tex]\dfrac{20+24}{84} *100[/tex]

= 0.5238095 × 100

= 52.83

To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%

Answer:

1. 98-103 (6th class)

2. 88

3. 5

4. 84

5. 52%

Step-by-step explanation:

Find attached the frequency table.

The class of exam scores falls between (1, 2, 3, 4, 5, or 6).

The exam score ranged from 68-103

1) The largest number of exam scores = 24

The largest number of exam scores is in the 6th class = 98 -103

Step 2 of 5:

The upper class limit is the higher number in an interval. Third class interval is 83-88

The upper class limit of the third class 88.

Step 3 of 5:

Class width = upper class limit - lower class limit

We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78

Class width = 78 - 73

Class width for the histogram = 5

Step 4 of 5:

The total of students that took the test = sum of all the frequency

= 0+4+8+12+16+20+24 = 84

The total of students that took the test = 84

Step 5 of 5:Find the percentage of students that scored higher than 95.5

Number of student that scored higher than 95.5 = 20 + 24 = 44

Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100

= (44/84) × 100 = 0.5238 × 100 = 52.38%

Percentage of students that scored higher than 95.5 = 52% (nearest percent)

Isabella averages 152 points per bowling game with a standard deviation of 14.5 points. Suppose Isabella's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(152,14.5)______.
If necessary, round to three decimal places.
Suppose Isabella scores 187 points in the game on Sunday. The z-score when x=187 is ___ The mean is _________
This z-score tells you that x = 187 is _________ standard deviations.

Answers

Answer:

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.

Step-by-step explanation:

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 pvalue, we get the probability that the value of the measure is greater than X.

In this question:

[tex]\mu = 152, \sigma = 14.5[/tex]

The z-score when x=187 is ...

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{187 - 152}{14.5}[/tex]

[tex]Z = 2.41[/tex]

The z-score when x=187 is 2.41. The mean is 187. This z-score tells you that x = 187 is 2.41 standard deviations above the mean.

According to a polling​ organization, 22​% of adults in a large region consider themselves to be liberal. A survey asked 200 respondents to disclose their political​ philosophy: Conservative,​ Liberal, Moderate. Treat the results of the survey as a random sample of adults in this region. Do the survey results suggest the proportion is higher than that reported by the polling​ organization? Use an alphaequals0.01 level of significance.

Answers

Answer:

No. There is not enough evidence to support the claim that the proportion of liberals is higher than that reported by the polling​ organization (P-value = 0.0366).

Step-by-step explanation:

The question is incomplete: there is no information about the results of the survey. We will assume that 55 of the subjects answer "liberal", and test the claim.

This is a hypothesis test for a proportion.

The claim is that the proportion of liberals is higher than that reported by the polling​ organization.

Then, the null and alternative hypothesis are:

[tex]H_0: \pi=0.22\\\\H_a:\pi>0.22[/tex]

The significance level is 0.01.

The sample has a size n=200.

The sample proportion is p=0.275.

[tex]p=X/n=55/200=0.275[/tex]

The standard error of the proportion is:

[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.22*0.78}{200}}\\\\\\ \sigma_p=\sqrt{0.000858}=0.029[/tex]

Then, we can calculate the z-statistic as:

[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.275-0.22-0.5/200}{0.029}=\dfrac{0.053}{0.029}=1.792[/tex]

This test is a right-tailed test, so the P-value for this test is calculated as:

[tex]\text{P-value}=P(z>1.792)=0.0366[/tex]

As the P-value (0.0366) is greater than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of liberals is higher than that reported by the polling​ organization.

A spinner has 10 equally sized sections, 8 of which are gray and 2 of which are blue. The spinner is spun twice. What is the probability that the first spin lands on gray and the second spin lands on blue ? Write your answer as a fraction in simplest form.

Answers

Answer:

4/25

Step-by-step explanation:

The probability the first spin lands on gray is 8/10 = 4/5.

The probability the second spin lands on blue is 2/10 = 1/5.

The probability of both events is 4/5 × 1/5 = 4/25.

Determine whether the data described below are qualitative or quantitative and explain why.

The area codes (such as 617 )of the telephones of survey respondents:

a. The data are quantitative because they consist of counts or measurements.
b. The data are quantitative because they don't measure or count anything.
c. The data are qualitative because they don't measure or count anything.
d. The data are qualitative because they consist of counts or measurements.

Answers

Answer:

c. The data are qualitative because they don't measure or count anything.

Step-by-step explanation:

In the case of the area codes, the value although is a number and follows some logic, it does not represent a quantity and any mathematical operation on it has no meaning. The number does not measure or count anything.

They have the same meaning as the name of the city or the area.

The sum of two fractions can always be written as a

Answers

Answer: decimal

Step-by-step explanation:

because i did this quiz

9. In 2002 the Georgia department of education reported a mean reading test score of 850 from Tattnall County Career Academy with a standard deviation of 50. The sample was taken from 100 11th grade students. Assuming the test scores are normally distributed, what is the standard error

Answers

Answer:

The standard error = 5

Step-by-step explanation:

Explanation:-

Given sample size 'n' = 100

Given mean reading test score μ =  850

Given standard deviation of the population 'σ' = 50

The standard error is determined by

 Standard error =   [tex]\frac{S.D}{\sqrt{n} }[/tex]

                    S.E = σ/√n

[tex]S.E = \frac{S.D}{\sqrt{n} } = \frac{50}{\sqrt{100} } = 5[/tex]

Final answer:-

The standard error ( S.E) = 5

1. In an arithmetic sequence, the first term is -2, the fourth term is 16, and the n-th term is 11,998

(a) Find the common difference d

(b) Find the value of n.​


pls help...

Answers

Answer:

see explanation

Step-by-step explanation:

The n th term of an arithmetic sequence is

[tex]a_{n}[/tex] = a₁ + (n - 1)d

(a)

Given a₁ = - 2 and a₄ = 16, then

a₁ + 3d = 16 , that is

- 2 + 3d = 16 ( add 2 to both sides )

3d = 18 ( divide both sides by 3 )

d = 6

--------------

(b)

Given

[tex]a_{n}[/tex] = 11998 , then

a₁ + (n - 1)d = 11998 , that is

- 2 + 6(n - 1) = 11998 ( add 2 to both sides )

6(n - 1) = 12000 ( divide both sides by 6 )

n - 1 = 2000 ( add 1 to both sides )

n = 2001

------------------

Which expression is equivalent to 5^10 times 5^5. 5^2 5^5 5^15 5^50

Answers

Answer:

5^15

Step-by-step explanation:

(5^10)(5^5)= 5^10+5= 5^15

Simplify the following expression:
-5[(x^3 + 1)(x + 4)]​

Answers

Answer:

[tex]-5x^{4} -20x^{3} -5x-20[/tex]

Step-by-step explanation:

[tex]-5[(x^{3} +1)(x+4)][/tex]

Use the FOIL method for the last two groups.

[tex]-5(x^{4} +4x^{3} +x+4)[/tex]

Now, distribute the -5 into each term.

[tex]-5x^{4} -20x^{3} -5x-20[/tex]

which of the following explains expressions are equivalent to - 5/6 /-1/3

Answers

Answer:

2.5

Step-by-step explanation:

(-5/6 ) / (-1/3)

multiply the numerator and denominator by the same number -3 gives:

(-5 * -3 /6 ) / (-1* -3/3)

(15/6 ) / (3/3)

(15/6 ) / 1

(15/6 )

12/6 + 3/6

2 3/6

2 1/2

2.5

Design and complete a frequency table for Belinda.
Belinda ask 20 people, how many hours of TV did you watch last week?

Here is the results
3,17,4,4,6,11,14,14,1,20,9,8,9,6,12,7,8,13,13,9.
Belinda wants to show these result in a frequency table.
She will use 4 equal groups.
The first group will start with 1 hour and the last group will end with 20 hours.

Answers

Answer:

Step-by-step explanation:

Since she will use 4 groups or class intervals, the the class width would be 20/4 = 5 hours

The class groups would be

1 to 5

5 to 10

10 to 15

15 to 20

The class mark for each class is the average of the minimum and maximum value of each class. Therefore, the class marks are

(1 + 5)/2 = 3

(5 + 10)/2 = 7.5

(10 + 15)/2 = 12.5

(15 + 20)/2 = 17.5

The frequency table would be

Class group Frequency

1 - 5 4

5 - 10 8

10 - 15 6

15 - 20 2

The total frequency is 4 + 8 + 6 + 2 = 20

For what value of the variable will the value of 7y−2 be ten more than the value of 2y?

Answers

Answer:

y=2.4

Step-by-step explanation:

7y-2=2y+10

7y-2y=10+2

5y=12

y=12/5=2.4

The average math SAT score is 511 with a standard deviation of 119. A particular high school claims that its students have unusually high math SAT scores. A random sample of 55 students from this school was​ selected, and the mean math SAT score was 528. Is the high school justified in its​ claim? Explain. ▼ No Yes ​, because the​ z-score ​( nothing​) is ▼ unusual not unusual since it ▼ does not lie lies within the range of a usual​ event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means. ​(Round to two decimal places as​ needed.)

Answers

Answer:

No, because the​ z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual​ event, namely within 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Unusual

If X is more than two standard deviations from the mean, x is considered unusual.

In this question:

[tex]\mu = 511, \sigma = 119, n = 55, s = \frac{119}{\sqrt{55}} = 16.046[/tex]

A random sample of 55 students from this school was​ selected, and the mean math SAT score was 528. Is the high school justified in its​ claim?

If Z is equal or greater than 2, the claim is justified.

Lets find Z.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{528 - 511}{16.046}[/tex]

[tex]Z = 1.06[/tex]

1.06 < 2, so 528 is not unusually high.

The answer is:

No, because the​ z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual​ event, namely within 2 standard deviations of the mean of the sample means.

The statement that could be made regarding the high school about the justification of its claim would be:

- No, because the​ z-score of Z = 1.06 is not unusual since it does not lie within the range of a usual​ event, namely within 2 standard deviations of the mean of the sample means.

Given that,

μ = 511

σ = 119

Sample(n) = 55

and

s = [tex]119/\sqrt{55}[/tex]

[tex]= 16.046[/tex]

As we know,

The claim of the high school could be valid and justified only when

[tex]Z > 2[/tex]

To find,

The value of Z

So,

[tex]Z = (X -[/tex] μ )/σ

by putting the values using Central Limit Theorem,

[tex]Z = (528 - 511)/16.046[/tex]

[tex]Z = 1.06[/tex]

Since [tex]Z < 2[/tex], the claim is not justified.

Learn more about "Standard Deviation" here:

brainly.com/question/12402189

Find the area of a triangle bounded by the y-axis, the line f(x)=9−2/3x, and the line perpendicular to f(x) that passes through the origin. Area =

Answers

Answer:

The area of the triangle is 18.70 sq.units.

Step-by-step explanation:

It is provided that a triangle is bounded by the y-axis, the line [tex]f(x)=y=9-\frac{2}{3}x[/tex].

The slope of the line is: [tex]m_{1}=-\frac{2}{3}[/tex]

A perpendicular line passes through the origin to the line f (x).

The slope of this perpendicular line is:[tex]m_{2}=-\frac{1}{m_{1}}=\frac{3}{2}[/tex]

The equation of perpendicular line passing through origin is:

[tex]y=\frac{3}{2}x[/tex]

Compute the intersecting point between the lines as follows:

[tex]y=9-\frac{2}{3}x\\\\\frac{3}{2}x=9-\frac{2}{3}x\\\\\frac{3}{2}x+\frac{2}{3}x=9\\\\\frac{13}{6}x=9\\\\x=\frac{54}{13}[/tex]

The value of y is:

[tex]y=\frac{3}{2}x=\frac{3}{2}\times\frac{54}{13}=\frac{81}{13}[/tex]

The intersecting point is [tex](\frac{54}{13},\ \frac{81}{13})[/tex].

The y-intercept of the line f (x) is, 9, i.e. the point is (0, 9).

So, the triangle is  bounded by the points:

(0, 0), (0, 9) and [tex](\frac{54}{13},\ \frac{81}{13})[/tex]

Consider the diagram attached.

Compute the area of the triangle as follows:

[tex]\text{Area}=\frac{1}{2}\times 9\times \frac{54}{13}=18.69231\approx 18.70[/tex]

Thus, the area of the triangle is 18.70 sq.units.

Which number is irrational

Answers

Answer:

Can you give the question. Can you post the picture. I can help solve. I will edit this answer once you have given the question/picture.

What’s the correct answer for this?

Answers

Answer:

C.

Step-by-step explanation:

Density = Mass / Volume

2.7 = 54 / V

V = 54 / 2.7

V = 20 cubic cm

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is randomly selected, find the probability that the class length is between 50.1 and 51.1 min. P(50.1 < X < 51.1) =

Answers

Answer:

P(50.1 < X < 51.1) = 0.5

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X between c and d is given by the following formula:

[tex]P(c < X < d) = \frac{d - c}{b - a}[/tex]

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.

This means that [tex]a = 50, b = 52[/tex]

So

[tex]P(50.1 < X < 51.1) = \frac{51.1 - 50.1}{52 - 50} = 0.5[/tex]

Given that d is the midpoint of line segment ab and k is the midpoint of line segment bc, which statement must be true? (May give brainliest)

Answers

Answer:

B is the midpoint of line segment AC

What else would need to be congruent to show that ABC=DEF by SAS?

Answers

Answer:

A

Step-by-step explanation:

Answer:

The answer here is A.

A) A is congruent to D.

A=

Step-by-step explanation:

AP E

a football team had 50 players at the start of the season, but then some players left the team. After that, the team had 42 players

Answers

Answer:

50 = p + 42

Step-by-step explanation:

The unknown part of this equation is the variable p, the number of people that left. So you want to add p to 42 and that will give you the total number of football players, which is 50. In order to get p, you need to get it by itself and make it equal something. Subtract 42 from both sides and you are stuck with 50-42 = p

p = 8

Answer:

50-p=42

Step-by-step explanation:

A commuter uses a bus and a train to get to work. The train is more than 5 minutes late 1/6 of the times they use it The bus is more than 5 minutes late 3/5 of the times they use it. What is the probability that both the bus and train will be more than 5 minutes late?

Answers

Answer:

10% probability that both the bus and train will be more than 5 minutes late

Step-by-step explanation:

Independent events:

If two events, A and B, are independent, we have that:

[tex]P(A \cap B) = P(A)*P(B)[/tex]

What is the probability that both the bus and train will be more than 5 minutes late?

The bus being more than 5 minutes late is independent of the train, and vice versa. So

Event A: Bus more than 5 minutes late

Event B: Train more than 5 minutes late

The train is more than 5 minutes late 1/6 of the times they use it

This means that [tex]P(B) = \frac{1}{6}[/tex]

The bus is more than 5 minutes late 3/5 of the times they use it.

This means that [tex]P(A) = \frac{3}{5}[/tex]

Then

[tex]P(A \cap B) = \frac{3}{5}*\frac{1}{6} = \frac{3}{30} = 0.1[/tex]

10% probability that both the bus and train will be more than 5 minutes late

Evaluate 1/2 + 1/2 ÷ 18​

Answers

Answer:

1/18

Step-by-step explanation:

First you would add 1/2 and 1/2 to get 1 then you would divide it by 18 to get 1/18

Answer:

1/18

Step-by-step explanation:

plz mark me brainliest.

HELP! Let f(x) = x + 1 and g(x)=1/x The graph of (fg)(x) is shown below.

Answers

Answer:

Step-by-step explanation:

all numbers except y = 1

because (f*g)(x) = 1+1/x

and 1/x cannot be equal to 0

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