10
Select the correct answer.
The given equation has been solved in the table.
Step
1
2
3₂
4
5
-
Statement
-7--7
7+7=-7+7
0
2
22-0
2=0
=
In which step was the subtraction property of equality applied?
O A. step 2
OB.
step 3
OC.
step 4
O D.
The subtraction property of equality was not applied to solve this equation.

The way to transform A to B is (1,-1).

We are given that;

The coordinates (-1,1),(-3,1),(-1,3) and (1,-1),(3,-1),(1,-3).

Now,

To transform the coordinates (-1,1),(-3,1),(-1,3) to (1,-1),(3,-1),(1,-3), you can apply a reflection about the origin. This transformation can be represented by the matrix [-1 0; 0 -1]. When you multiply this matrix with the original coordinates, you will get the transformed coordinates.

First, represent the point (-1,1) as a column vector:

[-1]

[ 1]

Then, multiply this vector by the reflection matrix [-1 0; 0 -1]:

[-1 0; 0 -1] * [-1] = [ 1]

[ 1]   [-1]

Therefore, by transformation the answer will be (1,-1).

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7 significant figures.

Zeros that come at the start do not count as significant figures.

Eg. 0.0001 would only be 1 significant figure.

Zeros that come after is counted as a significant figure.

Eg. 2.20 Is three significant figures.

Hello!

[tex]3x^{2} + 7x-12x - 34 - 2x^{2} + 10\\\\3x^{2}- 2x^{2} + 7x-12x - 34 + 10\\\\x^{2} - 5x - 24\\\\x^{2} + 3x - 8x - 24\\\\(x^{2} + 3x) + (-8x - 24)\\\\x(x + 3) - 8(x + 3)\\\\\boxed{(x - 8)(x+3)}[/tex]

Answer:

When you cut the sample size in half while making a statistical inference about the mean of a normally distributed population, the effect on the margin of error depends on the relationship between the sample size and the margin of error. Generally, the margin of error is inversely proportional to the square root of the sample size.

So, if you reduce the sample size by half, it means you are taking a smaller sample, which will result in a larger margin of error. In other words, the margin of error is multiplied by a factor greater than 1.

Among the given options, the correct answer is:

D. The margin of error is multiplied by 2.

This option correctly reflects the relationship between reducing the sample size by half and the resulting increase in the margin of error.

[tex]a^2+ab=a(a+b)\\ab+b^2=b(a+b)[/tex]

Therefore

[tex]\text{hcf}(a^2+ab,ab+b^2)=a+b[/tex]

Work Shown:

LCM = (product of two numbers)/(HCF of the two numbers)

LCM = 1536/16

LCM = 96

Answer: [tex]=\frac{2x}{x-3}[/tex]  

Step-by-step explanation:

 [tex]\frac{2x^{2} +10x}{x^{2} + 2x-15}[/tex]                        >Take out GCF(greatest common factor) on top

                                       >factor the bottom,  find 2 numbers that mulitply to

                                         "c" term, -15 and adds to the "b" term, +2

                                        >+5 and -3 mulitpy to -15 and add to +2

                                        > put +5 and -3 into factored form on bottom

[tex]=\frac{2x(x+5)}{(x-3)(x+5)}[/tex]                      >reduce the x+5 on top and bottom

[tex]=\frac{2x}{x-3}[/tex]                                >This is simplified

Mr. Jones in Johannesburg is billed R9,767.12 and Ms. Brown in Cape Town is billed R680.24 for their respective properties.

From the provided information for Cape Town's sanitation tariffs:

0-4.2 kl: R16.03 per kl

Since 4.1 ke is equivalent to 4100 liters, which is less than 4.2 kl, we can use the tariff rate for the 0-4.2 kl range.

Cost for 4.1 ke of sanitation

= 4.1 ke x R16.03 per kl

= 4100 liters x R16.03 per kl

= R65.83

For Mr. Jones and Ms. Brown, who own properties with an area of 550 m² each and were billed for 22 kl of sanitation.

we need to determine the applicable tariff rate based on the property size and calculate the cost.

In Johannesburg, based on the provided information, the tariff rate for properties larger than 300 m² to 1,000 m² is R443.96.

Cost of sanitation for Mr. Jones in Johannesburg:

= 22 kl x R443.96 per kl

= R9,767.12

Cost of sanitation for Ms. Brown in Cape Town:

= 22 kl x R30.92 per kl = R680.24

Therefore, Mr. Jones in Johannesburg is billed R9,767.12 and Ms. Brown in Cape Town is billed R680.24 for their respective properties.

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The value of variables in the figure is,

⇒ x = 80

⇒ y = 70

We have to given that;

By using given figure we have to find the value of each variable.

Now, We can formulate;

⇒ 55 = 1/2 (180 - y)

Solve for y;

⇒ 55 × 2 = 180 - y

⇒ 110 = 180 - y

⇒ y = 180 - 110

⇒ y = 70

And, The value of x is,

⇒ x = 180 - (60 + 40)

⇒ x = 180 - 100

⇒ x = 80

Thus, The value of variables in the figure is,

⇒ x = 80

⇒ y = 70

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Answer:  4, 14

Step-by-step explanation:

Bring everything over to other side that is not in the absolute value:

3 |x - 9| + 5 = 20                  >Subtract 5 from both sides

3 |x - 9| = 15                         >Divide both sides by 3

|x - 9| = 5                              >Create a positive and negative version to

                                                 drop the absolute value

x - 9 = 5                        x - 9 = -5

x= 14                                    x = 4

Answer:

3

Step-by-step explanation:

To find the z-score for a data value of 155 in a normal distribution with a mean of 137 and a standard deviation of 6, you can use the formula:

z = (x - μ) / σ

where:

x is the data value,

μ is the mean, and

σ is the standard deviation.

Plugging in the values, we have:

z = (155 - 137) / 6

Calculating this expression:

z = 18 / 6 = 3

It is generally appropriate to calculate a probability about the difference in sample means when the sample sizes are large (typically n ≥ 30) and the populations are approximately normally distributed.

To determine if it is appropriate to calculate a probability about the difference in sample means, consider the assumptions and conditions for conducting a hypothesis test or constructing a confidence interval. The appropriateness of calculating a probability about the difference in sample means depends on the following factors:

1) Both population shapes are unknown. n1 = 50 and n2 = 100:

- It is generally appropriate to calculate a probability about the difference in sample means when the sample sizes are large (typically n ≥ 30) and the populations are approximately normally distributed. Since the population shapes are unknown in this case, it is difficult to assess this condition. However, the large sample sizes (n1 = 50 and n2 = 100) may suggest that it is reasonable to approximate the population distributions as normal. Therefore, calculating a probability about the difference in sample means could be considered.

2) Population 1 is skewed right and population 2 is approximately Normal. n1 = 50 and n2 = 10:

- In this case, the assumption of approximately normally distributed populations is violated for population 1, which is skewed right. When the population distributions are not approximately normal, it may not be appropriate to calculate a probability about the difference in sample means. The small sample size for population 2 (n2 = 10) may also limit the accuracy of any inference made based on this sample.

3) Both populations are skewed right. n1 = 5 and n2 = 10:

- Similar to the previous case, the assumption of approximately normally distributed populations is violated for both populations. Additionally, the small sample sizes (n1 = 5 and n2 = 10) may not provide sufficient information for reliable inferences. Therefore, it is generally not appropriate to calculate a probability about the difference in sample means in this situation.

4) Population 1 is skewed right and population 2 is approximately Normal. n1 = 10 and n2 = 50:

- Similar to case 2, the assumption of approximately normally distributed populations is violated for population 1, which is skewed right. In this case, the sample size for population 1 (n1 = 10) is also small, which may limit the accuracy of any inference made based on this sample. The larger sample size for population 2 (n2 = 50) might make it more reasonable to approximate the population distribution as normal. However, the violation of the assumption for population 1 suggests caution when interpreting the results. It is not generally appropriate to calculate a probability about the difference in sample means in this situation.

5) Both populations have unknown shapes. n1 = 50 and n2 = 100:

- Similar to case 1, the population shapes are unknown. However, the large sample sizes (n1 = 50 and n2 = 100) might suggest that it is reasonable to approximate the population distributions as normal. As mentioned before, calculating a probability about the difference in sample means could be considered in this case.

6) Both populations are skewed left. n1 = 5 and n2 = 40:

The assumption of approximately normally distributed populations is violated for both populations, as they are skewed left. Additionally, the small sample sizes (n1 = 5 and n2 = 40) may not provide sufficient information for reliable inferences. Therefore, it is generally not appropriate to calculate a probability about the difference in sample means in this situation.

Summarily, it is generally appropriate to calculate a probability about the difference in sample means when the sample sizes are large (typically n ≥ 30) and the populations are approximately normally distributed. However, when the population distributions are not approximately normal or when the sample sizes are small, it is generally not appropriate to calculate a probability about the difference in sample means.

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Hello :)

P(A) = 1/2

P(B) = 5 or greater = 5,6 => 2/6 = 1/3

P(A and B) = 1/2 x 1/3 = 1x1/2x3 = 1/6 ≈ 0.17

the answer is 0.17

Answer:

[tex]a_n=3n-16[/tex]

Step-by-step explanation:

[tex]a_n=a_1+(n-1)d\\\\a_8=a_1+(8-1)d\rightarrow 8=a_1+7d\\a_{20}=a_1+(20-1)d\rightarrow 44=a_1+19d[/tex]

[tex]-36=-12d\\3=d[/tex]

[tex]8=a_1+7(3)\\8=a_1+21\\-13=a_1[/tex]

[tex]a_n=-13+(n-1)(3)\\a_n=-13+3n-3\\a_n=3n-16[/tex]

Option B has the highest overall cost by $64.50.

To determine the option that would result in the highest overall cost for the employee

we need to compare the costs of both options based on the estimated prescription drug expenses.

Option A:

Monthly premium: $94

Prescription coverage: 80%

Option B:

Monthly premium: $42

Prescription coverage: 75% for the first $500, 85% for costs over $500

Let's calculate the costs for each option:

Option A:

Total prescription drug cost: $1,250

Employee's share (20%): 20% × $1,250 = $250

Monthly premium: $94

Total cost for Option A: $250 + $94 = $344

Option B:

Total prescription drug cost: $1,250

Employee's share for the first $500 (25%): 25% × $500 = $125

Employee's share for costs over $500 (15%): 15% × ($1,250 - $500) = $112.50

Monthly premium: $42

Total cost for Option B: $125 + $112.50 + $42 = $279.50

Comparing the total costs for each option, we see that Option B has a lower overall cost for the employee.

Hence, Option B has the highest overall cost by $64.50.

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Answer:c

Step-by-step explanation:

During the Christmas holidays, there are 1,053 more passengers who are late for their planes each day compared to a normal day.

Calculating the difference between the quantity of late passengers on a typical day and the quantity of late passengers on holidays is necessary.

1,835 travellers were late over the Christmas break.

On an average day, there are 782 travelers who are late.

Difference = Number of late passengers during the Christmas holidays - Number of late passengers on a normal day

Difference = 1,835 - 782

Difference = 1,053

Therefore, during the Christmas holidays, there are 1,053 more passengers who are late for their planes each day compared to a normal day.

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Answer:

Step-by-step explanation:

During the Christmas holidays, there is an increase of 1,053 passengers who are late for their planes each day compared to the average daily number of 782 passengers who are late. This surge in late passengers during the holiday period contributes to delays in 14 planes.

The higher volume of travelers, combined with potential factors like weather conditions, congestion, and heightened travel demand, likely leads to a greater number of individuals encountering delays and difficulties reaching their departure gates on time.

Hope it helps! :)

Hello!

Solutions = -3 and 1.5

Answer:

664 Square yards

Step-by-step explanation:

Surface area= 2lw+2lh+2hw

=2(17×6)+2(17×10)+2(10×6)

=2(102)+2(340)+2(120)

=204+340+120

=664 Square yards

Answer:

119

Step-by-step explanation:

If you want to find the base area of a rectangular prism, then all you need is length and width only.

[tex] \boxed{Volume = length \times width \times height} [/tex]

You don't need height. So, this is a must be easy.

[tex]\blue{\small{\mathfrak{That's \: it. \: Thanks \::)}}} [/tex]

A construction of the box-and-whisker plot representing the data set is shown below.

In Mathematics and Statistics, a box-and-whisker plot and it can be defined as a type of chart that can be used to graphically or visually represent the five-number summary of a data set with respect to locality, skewness, and spread.

Based on the data (information) provided above, the five-number summary for the given data set include the following:

In conclusion, we can logically deduce that the maximum number is 100 while the minimum number is 35, and the median is equal to 88.

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Answer:

The optimal size of each production run is 180 pounds, and the total annual inventory cost is $832,825.

Step-by-step explanation:

To determine the optimal size and total annual inventory cost, we need to consider the production capacity, demand, setup costs, and carrying costs.

Given:

Production capacity per day = 250 pounds

Demand per day = 180 pounds

Setup cost = $125

Carrying cost per pound per year = $12

First, let's calculate the optimal production size. We want to produce enough cheese to meet the demand without exceeding the production capacity.

Optimal production size per day = Minimum(Production capacity, Demand)

Optimal production size per day = Minimum(250 pounds, 180 pounds) = 180 pounds

Next, let's calculate the total annual inventory cost. This cost includes both the setup cost and the carrying cost.

Total annual inventory cost = (Setup cost * Number of setups per year) + (Carrying cost per pound * Optimal production size * Number of days in a year)

Number of setups per year = Number of production runs per year

Number of production runs per year = Total days in a year / Days per production run

Assuming a year has 365 days and each production run takes one day:

Number of production runs per year = 365 days / 1 day = 365 runs

Total annual inventory cost = ($125 * 365) + ($12 * 180 pounds * 365)

Total annual inventory cost = $45,625 + $787,200

Total annual inventory cost = $832,825

To prove ∠AOW = 45°

Given,

∠WOZ = 90°

∠ZOB = 45°

Now,

∠XOA +∠AOW = 90°............(1)

∠XOA = ∠ZOB ( Vertically opposite angle  )

∠XOA = 45°

Substitute in (1),

45° +  45° = 90°

Now,

∠WOX = ∠XOA +∠WOA

So,

∠WOA + ∠WOZ + ∠ZOB = 180°( Linear pair )

∠WOA + 90° + 45° = 180°

∠WOA = 45°

Hence proved.

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Answer:

False

Step-by-step explanation:

1. The inverse function should have c(n) isolated
2. When finding the inverse of a function, the variables c(n) and n are interchanged (and then c(n) is isolated).
It would look like this --->c(n)=50+4n--->n=50+4(c(n)) ---> c(n)=(n-50)/4

Answer:

D 0.89275

Step-by-step explanation:

The 95% confidence interval for the mean number of delivery orders on a Friday night is 36 to 60 orders.

The confidence interval for the mean number of delivery orders on a Friday night is constructed assuming a 95% confidence level.

Given data:

Sample mean (x) = 48 orders

The margin of error (E) = 12 orders

To calculate the confidence interval, we'll use the formula:

Confidence Interval = x ± E

Substituting the values:

Confidence Interval = 48 ± 12

The lower limit of the confidence interval:

Lower Limit = 48 - 12 = 36

The upper limit of the confidence interval:

Upper Limit = 48 + 12 = 60

Therefore, the 95% confidence interval for the mean number of delivery orders on a Friday night is 36 to 60 orders.

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Complete question:

In Boston, A Popular Restaurant Recently Added A Delivery Option. In Reviewing The Sales For This New Delivery Service, The Restaurant Found The Mean Delivery Orders In A Random Sample Of Friday Nights Was X¯=48 Orders, With A Margin Of Error Of 12 Orders. Construct A Confidence Interval For The Mean Number Of Delivery Orders On A Friday Night.