To the nearest whole number, what is the value of m? 13 14 43 45

Answer:

136

Step-by-step explanation:

Angle G and Angle H are corresponding angles so they equal themselves;

6x + 34 = 8x

Subtract 6x from both sides;

34 = 2x

Divide both sides by 2;

x = 17

Angle H;

8x = 8(17) = 136

Answer:b the probabilty that it contains nuts and is white

Step-by-step explanation:im sorry if this is wrong but its the only one that makes sense to me

Conditional probability is represented by probability that the candy is blue

The concept of the conditional probability formula is one of the quintessential concepts in probability theory. The conditional probability formula gives the measure of the probability of an event, say B given that another event, say A has occurred.  

The Bayes' theorem is used to determine the conditional probability of event A, given that event B has occurred, by knowing the conditional probability of event B, given that event A has occurred, also the individual probabilities of events A and B.

: In case P(B)=0, the conditional probability of P(A | B) is undefined.  (the event B did not occur)

Given:

Some red, white, and blue candies were placed in a bowl.

Some contain nuts, and some do not

Here we are only focusing on the red candy which shows we have reduce the sample space.

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The resistance of a 20-ft wire with a diameter of 0.1-in is 0.08ω

The variation is given as:

Resistance varies directly as its length and inversely as square of diameter

Mathematically, this is represented as

R =kL/D^2

Where k represents the constant of variation.

When L = 50, D = 0.4 in and R = 0.0125, we have

0.0125 =k * 50/0.4^2

This gives

k = 0.4^2 * 0.0125/50

Evaluate

k = 0.00004

Substitute k = 0.00004 in R =kL/D^2

R = 0.00004L/D^2

When L = 20-ft wire and D = 0.1-in, we have:

R = 0.00004 * 20/(0.1)^2

Evaluate

R = 0.08

Hence. the resistance of a 20-ft wire with a diameter of 0.1-in is 0.08ω

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Confidence intervals provide a range of values that likely contain the true population parameter. A two-tailed z-test showed significant differences in mean scores for the verbal exam and years of education compared to the statewide average.

1a) Mean, x = $478.23, Standard deviation, s = $15.78, Sample size, n = 429, Confidence level, C.L = 99%. As the sample size is greater than 30, we can use z-test.\(\[Z = \frac{\left( {x - \mu } \right)}{\frac{s}{\sqrt n }}\]\Rightarrow Z = \frac{\left( {478.23 - \mu } \right)}{\frac{{15.78}}{{\sqrt {429} }}}\]\)

At 99% confidence level, level of significance (α) = 1% = 0.01. So, α/2 = 0.005 as the test is two-tailed. So, the level of confidence (C.L) = 99% = (1 - α) × 100% = 99%. At a 99% confidence level, the critical value, Zc = 2.58 (approx)

Calculation\(:\[2.58 = \frac{\left( {478.23 - \mu } \right)}{\frac{{15.78}}{{\sqrt {429} }}}\]\Rightarrow \frac{{2.58 \times \frac{{15.78}}{{\sqrt {429} }}}}{{478.23 - \mu }} = 1\]\). Taking reciprocal on both sides, we get,\(\[\frac{{478.23 - \mu }}{{2.58 \times \frac{{15.78}}{{\sqrt {429} }}}} = 1\]\Rightarrow 478.23 - \mu = 2.58 \times \frac{{15.78}}{{\sqrt {429} }}\]\Rightarrow \mu = 478.23 - 2.58 \times \frac{{15.78}}{{\sqrt {429} }}\]\).

The lower limit of the confidence interval, \(LCL \[= \mu - Zc\left( {\frac{s}{{\sqrt n }}} \right) = 478.23 - 2.58\left( {\frac{{15.78}}{{\sqrt {429} }}} \right)\]\)

The upper limit of the confidence interval, \(UCL \[= \mu + Zc\left( {\frac{s}{{\sqrt n }}} \right) = 478.23 + 2.58\left( {\frac{{15.78}}{{\sqrt {429} }}} \right)\]\). Thus, the 99% confidence interval for the mean of the population of textbook spending is $475.29 to $481.17.1

b) Mean, x = 2.8, Standard deviation, s = 1.0, Sample size, n = 429, Confidence level, C.L = 99%. As the sample size is greater than 30, we can use z-test.

\(\[Z = \frac{\left( {x - \mu } \right)}{\frac{s}{\sqrt n }}\]\Rightarrow Z = \frac{\left( {2.8 - \mu } \right)}{\frac{1}{{\sqrt {429} }}}\].\) At 99% confidence level, level of significance (α) = 1% = 0.01. So, α/2 = 0.005 as the test is two-tailed. So, the level of confidence (C.L) = 99% = (1 - α) × 100% = 99%. At a 99% confidence level, the critical value, Zc = 2.58 (approx)

Calculation:\(\[2.58 = \frac{\left( {2.8 - \mu } \right)}{\frac{1}{{\sqrt {429} }}}} \Rightarrow \frac{{2.58 \times \frac{1}{{\sqrt {429} }}}}{{2.8 - \mu }} = 1\]\). Taking reciprocal on both sides, we get,\(\[\frac{{2.8 - \mu }}{{2.58 \times \frac{1}{{\sqrt {429} }}}}\]\). Thus, the 99% confidence interval for the mean of the population of missed classes is 2.68 to 2.92.

2) Mean, x = 502, Standard deviation, s = 95, Sample size, n = 137, Level of significance (α) = 0.05. At a 5% level of significance, we can assume the distribution to be normal as n > 30. Critical value of Z at α/2 = 0.025 level of significance = 1.96 (approx)

Calculation\(:\[Z = \frac{{\left( {x - \mu } \right)}}{{\frac{s}{{\sqrt n }}}} = \frac{{\left( {502 - 453} \right)}}{{\frac{{95}}{{\sqrt {137} }}}} = 4.055\]\). As Z > Zc, we can reject the null hypothesis. Therefore, there is a statistically significant difference in the mean score of the verbal portion of the college entrance exam.

3) Mean, x = 12.7, Standard deviation, s = 1.7, Sample size, n = 423, Level of significance (α) = 0.05, Population mean, μ = 12.2. At a 5% level of significance, we can assume the distribution to be normal as n > 30. Z-test is used to test the difference in means.

\(\[Z = \frac{{\left( {\rm{x}} - \rm{y}}} \right) - \left( {\rm{\mu }}_\rm{x}} - \rm{\mu }}_\rm{y}}} \\\right)}}{{\sqrt \frac{{\rm{s}}_{\rm{x}}^2}}{\rm{n}} + \frac{{\rm{s}}_\rm{y}}^2}}\rm{m}}} }\]\), where x is the sample, y is the population, σ is the standard deviation, and n is the sample size.

Calculation:\(\[Z = \frac{{\left( {12.7 - 12.2} \right) - 0}}{{\sqrt {\frac{{1.7^2}}{423} + \frac{{0^2}}{n}}} } = 5.933\]\). As Z > Zc, we can reject the null hypothesis. Therefore, it is significantly different from the statewide average of 12.2 years.

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The exact function values are:

(a) sin 2t = -4/5

(b) cos 2t = -3/5

(c) tan 2t = 4/3

In Quadrant II, the sine of the t is positive and the cosine of the t is negative. After Using the double angle formulas, we can easily find the values of sin 2t, cos 2t, and tan 2t.

(a) sin 2t = 2sin t cos t = 2(-2/5)(-3/5) = -4/5

(b) cos 2t = cos² t - sin² t = (-3/5)² - (-2/5)² = -9/25 - 4/25 = -3/5

(c) tan 2t = (2tan t)/(1-² t) = (2(-2/5))/(1-(-2/5)²) = 4/3.

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True. Hexadecimal numbers are written using the base 16 number system, where digits range from 0 to 9 and A to F. They are commonly used in computer systems for concise representation and easy conversion to binary.

In the hexadecimal number system, there are 16 symbols used to represent values, namely 0-9 and A-F. Each digit in a hexadecimal number represents a multiple of a power of 16.

The symbols 0-9 represent the values 0-9, respectively. The symbols A-F represent the values 10-15, respectively, where A represents 10, B represents 11, C represents 12, D represents 13, E represents 14, and F represents 15.

For example, the hexadecimal number "3F" represents the value (3 * 16^1) + (15 * 16^0) = 48 + 15 = 63 in decimal.

Similarly, the hexadecimal number "AB8" represents the value (10 * 16^2) + (11 * 16^1) + (8 * 16^0) = 2560 + 176 + 8 = 2744 in decimal.

Hexadecimal numbers are commonly used in computer systems, as they provide a convenient way to represent large binary numbers concisely. Each hexadecimal digit corresponds to a four-bit binary number, allowing for easy conversion between binary and hexadecimal representations.

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Answer:

so like would that be a cute

Step-by-step explanation:

because it's under 90 degrees

Answer:

b or d

Step-by-step explanation:

Answer:

-7

Step-by-step explanation:

f(10) = -2* 10 +13

= -20 + 13

= -7

Answer:

$50

Step-by-step explanation:

Price of the cell phone bought p = $40

% of markup in price = 25%

Selling price of the cell phone = p + (%)p

Selling price of the cell phone = P + (0.25)p

Selling price of the cell phone = p + 0.25p

Selling price of the cell phone = 1.25p

Selling price of the cell phone = 1.25 * $40

Selling price of the cell phone = $50

David takes a car to get to the airport in time for his trip. 210 miles distance is the airport from his house in miles.

Given that,

David takes a car to get to the airport in time for his trip. He travels 35 miles in the first hour but understands that if he keeps going at this rate, he will be one hour late. For the final 15 miles to the airport, he speeds up by 15 mph and gets there 30 minutes early.

We have to find how far is the airport from his house in miles.

Distance traveled after 1 hour

Let us take the distance traveled after first 1 hour as y

Distance/speed = time

After 1 hour his speed = (35 + 15) = 50 mph

Total time count of 1 hour and 30 mins is 1.5 hr

y/50 + 1.5 = y/35

Multiply through by350

7y + 525 = 10y

3y = 525

y = 525/3

y = 175 miles

In the first 1 hour, at 35 mph, distance covered = 35 miles

Distance from his home to airport = 175 miles + 35 miles = 210 miles

Therefore, the distance between David's home and the airport is 210 miles.

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The correct matching of the items of math in the columns is:

A composite number is a number that possesses additional factors besides 1 and its own value. The smallest number that can be divided evenly by all the denominators given is known as the least common denominator.

Non-numeric criteria are used to group data in categorical data. Fractions that have identical numerical values are referred to as equivalent fractions. When a line segment passes through the center of a circle and joins two points on its circumference, it is known as the circle's diameter.

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The Solomon four-group design includes two control groups: one that is not pretested and does not receive treatment, and another that is not pretested but receives treatment.

In the Solomon four-group design, there are two treatment groups and two control groups. The purpose of this design is to examine the interaction effect between pretesting and treatment.

The first control group does not receive any treatment, while the second control group also does not receive treatment but is pretested. These two control groups help to measure the impact of pretesting on the dependent variable.

The two treatment groups receive the treatment being studied, with one group being pretested and the other group not being pretested. By comparing the pretested and non-pretested treatment groups, researchers can determine if there is an interaction effect between pretesting and the treatment.

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we have that

50%=50/100=0.50

Multiply the total tickets sold by the factor of 0.50

so

94*0.50=47 tickets

Answer:

The british

Step-by-step explanation:

Answer:

The Britsh

Step-by-step explanation:

Formal colonialism first came to the region we today call Ghana in 1874, and British rule spread through the region into the early twentieth century. The British called the territory the “Gold Coast Colony."

Hence, the margin of error for this 95% confidence interval of a population proportion is 5.4%.

The margin of error for a confidence interval is the distance between the sample statistic (in this case, the sample proportion) and the confidence interval limits. To find the margin of error, we can use the formula:

Margin of error = (upper limit - lower limit) / 2

In this case, the lower limit of the 95% confidence interval is 64.5% and the upper limit is 75.3%. So the margin of error is:

Margin of error = (75.3% - 64.5%) / 2

Margin of error = 10.8% / 2

Margin of error = 5.4%

Therefore, the margin of error for this 95% confidence interval of a population proportion is 5.4%.

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Answer:

456

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

Answer:

The set of all whole numbers

Step-by-step explanation:

Real numbers, integers, and rational numbers can all contain negative numbers. Whole numbers include numbers from 0 and up.

Hope this helps!

Have a nice day

14

Step-by-step explanation:

I'm pretty sure it's 14

Answer: y=-5x-1

Step-by-step explanation:

The two given points are (-1,4) and (1.-6)

We can then use the slope formula (Y2-Y1)/(X2-X1) to find the slope

4-(-6)/ -1-1 = 10/-2 = -5

Slope is -5

We then plug the slope in a set of points to find the y intercept

-5*-1+x = 4

5+x=4

x=-1

Then just add these two together to form the equation of y=mx+b

y=-5x-1

Answer:

x= -7/5

y=-6/5

one solution I believe. Hope this helps!

Answer: 1. (a) is 50 cents, (b) is 75 cents, and (c) is 50 cents.

2. It weighed 2-3 (2 to 3) ounces.

3. If we're being honest, I forgot how to do this type problem but the other two I'm sure of.

Step-by-step explanation:

Answer:

= or >14

Step-by-step explanation:

x needs to be equal to or greater than 14

Answer:

-14

Step-by-step explanation:

4n - 6 , n= -2

= 4* (-2) - 6

= -8 -6

= - 14

Answer:

import java.util.Scanner;

public class Main {

  public static void main(String[] args) {

      Scanner input = new Scanner(System.in);

      int n = input.nextInt();

      int l = 1;

      for(int i=1; i <= n; i++){

          for(int j=1; j<= l; j++){

              System.out.print("*");

          }

          System.out.println();

          l = l + 2;

      }

      l = l - 4;

      for(int i=1; i < n; i++){

          for(int j=1; j <= l; j++){

              System.out.print("*");

          }

          System.out.println();

          l = l-2;

Step-by-step explanation:

To begin, utilize a Scanner object to receive an integer as input (Line 5-6).

To print the upper section of the diamond form, make the first set of inner loops (Line 8-14). The variable l specifies the maximum number of asterisks that can be written in each row.

Create another inner loop to print the diamond shape's lower half (Line 17-23). The same variable l is used to specify the maximum number of asterisks to appear in each row.

To find the solution to the differential equation dz/dt=7te^(3z) that passes through the origin, we can use separation of variables. The solution to the differential equation dz/dt = 7te^(3z) that passes through the origin is z(t) = (-1/3)ln((-3/7)t^2 - 1/3).

First, we write the equation as:
1/e^(3z) dz/dt = 7t
Then, we integrate both sides with respect to t and z:
∫1/e^(3z) dz = ∫7t dt
Solving the integral on the left side, we get:
-1/3 e^(-3z) = 7t^2/2 + C
where C is the constant of integration.
To find the value of C that satisfies the condition of passing through the origin, we substitute t=0 and z=0 into the equation:
-1/3 e^(0) = 7(0)^2/2 + C
-1/3 = C
Therefore, the solution to the differential equation dz/dt=7te^(3z) that passes through the origin is:

-1/3 e^(-3z) = 7t^2/2 - 1/3
To find the solution to the differential equation dz/dt = 7te^(3z) that passes through the origin, follow these steps:
1. Identify the given differential equation: dz/dt = 7te^(3z).
2. Notice that this is a first-order, separable differential equation. Separate the variables by dividing both sides by e^(3z) and multiplying both sides by dt:
  (1/e^(3z))dz = 7t dt
3. Integrate both sides of the equation with respect to their respective variables:
  ∫(1/e^(3z))dz = ∫7t dt
4. Evaluate the integrals:
  (-1/3)e^(-3z) = (7/2)t^2 + C
5. Solve for z:
  e^(-3z) = (-3/7)t^2 + C*(-3)
6. Apply the initial condition that the solution passes through the origin (t = 0, z = 0):
  e^(0) = (-3/7)(0)^2 + C*(-3)
  1 = 0 + C*(-3)
7. Solve for C:
  C = -1/3
8. Plug the value of C back into the equation for z:
  e^(-3z) = (-3/7)t^2 - 1/3
9. Finally, to find the solution in terms of z(t), take the natural logarithm of both sides:
  -3z = ln((-3/7)t^2 - 1/3)
10. Isolate z:
  z(t) = (-1/3)ln((-3/7)t^2 - 1/3)

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