WILL GIVE BRAINLIST On a coordinate plane, 2 quadrilaterals are shown. The first figure has points A (negative 2, 1), B (negative 4, 1), C (negative 4, 5), and D (negative 2, 4). Figure 2 has points A prime (2, 1), B prime (4, 1), C prime (4, 5), and D prime (2, 4). What is the rule for the reflection? rx-axis(x, y) → (–x, y) ry-axis(x, y) → (–x, y) rx-axis(x, y) → (x, –y) ry-axis(x, y) → (x, –y)

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]

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:

Answer:

4/11

Step-by-step explanation:

There are 11 marbles in total, if 4 of them are white, then you have a 4/11 chance of getting a white marble.

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.

Answer:

y = 12 + 6x

y = 12x

Step-by-step explanation:

From the information provided, the following equations are derived:

y = 12 + 6x    ------- Eqn 1

y = 12x          ------- Eqn 2

Since Eqns 1 and 2 have the same subject, we equate them to solve for x. We have:

12x = 12 + 6x

Putting like terms together, we have:

12x - 6x = 12 ⇒ (12 - 6)x = 12

6x = 12 ⇒ x = 2

x = 2

Substitute x into Eqn 1 or 2

Eqn 1

y = 12 + 6x

y = 12 + 6(2) = 12 + 12

y = 24

Eqn 2

y = 12x

y = 12(2)

y = 24

It means that it will take Ms. Ironperson and Mr. Thoro 2 days apiece to produce the same number of posters at the current rate (which is 24 posters). Both Ms. Ironperson and Mr. Thoro will individually take 2 days to produce 24 Avenger posters apiece.

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

Answer:

c

Step-by-step explanation:

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.

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

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)

Answer:

[tex] E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8[/tex]

Now we can find the second moment with this formula:

[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And replacing we got:

[tex] E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4[/tex]

The variance would be given by:

[tex] Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96[/tex]

And the deviation would be:

[tex] Sd(X) =\sqrt{6.96}= 2.638[/tex]

Step-by-step explanation:

For this case we have the following distribution given:

X      1      2      7

P(X) 1/5  2/5   2/5

We need to begin finding the mean with this formula:

[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

And replacing we got:

[tex] E(X) =1 *\frac{1}{5} +2 *\frac{2}{5} +7*\frac{2}{5}= 3.8[/tex]

Now we can find the second moment with this formula:

[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And replacing we got:

[tex] E(X^2) =1^2 *\frac{1}{5} +2^2 *\frac{2}{5} +7^2*\frac{2}{5}= 21.4[/tex]

The variance would be given by:

[tex] Var(X) =E(X^2) -[E(X)]^2 = 21.4 -[3.8]^2 = 6.96[/tex]

And the deviation would be:

[tex] Sd(X) =\sqrt{6.96}= 2.638[/tex]

Answer:

Bailey: 16%

Coco: 28%

Ginger: 32%

Ruby: 24%

I hope this helps!

Answer:

C.

Step-by-step explanation:

11. There is no flaw since step 1 is given, and there is the right reason for step 2.

1. Molly is using inductive reasoning because she is collecting the data to make a conjecture.

11. There is no flaw.

Step-by-step explanation:

Find the lowest common multiple of 15 and 12.

Which is 60.

15×4=60 so 32x4=£1.28

12x5=60 so 43x5=£2.15

2.15+1.28= £3.43

Answer:

5^15

Step-by-step explanation:

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

Answer:

a. P(X=50)= 0.36

b. P(X≤75) =  0.9

c. P(X>50)=  0.48

d. P(X<100) = 0.9

Step-by-step explanation:

The given data is

x                      25        50       75        100          Total

P(x)                0.16       0.36    0.38    0.10          1.00

Where X is the variable and P(X) = probabililty of that variable.

From the above

a. P(X=50)= 0.36

We add the probabilities of the variable below and equal to 75

b. P(X≤75) = 0.16+ 0.36+ 0.38= 0.9

We find the probability of the variable greater than 50 and add it.

c. P(X>50)= 0.38+0.10=  0.48

It can be calculated in two ways. One is to subtract the probability of 100 from total probability of 1. And the other is to add the probabilities of all the variables less than 100 . Both would give the same answer.

d. P(X<100)= 1- P(X=100)= 1-0.1= 0.9

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

Answer:

C.

Step-by-step explanation:

Density = Mass / Volume

2.7 = 54 / V

V = 54 / 2.7

V = 20 cubic cm

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

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

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.

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.

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

Answer:

$976

Step-by-step explanation:

Straight time pay= $15.25(hourly rate) × 40(hours worked)= $610

Overtime Rate = 15.25×2= $30.50

Overtime Pay= $30.5 × 12 (Hours worked overtime)= $366

Total Pay= Basic wage + Overtime Wage = $976

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

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]

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

Answer:

B is the midpoint of line segment AC

Answer:

y=2.4

Step-by-step explanation:

7y-2=2y+10

7y-2y=10+2

5y=12

y=12/5=2.4