A bakery sold 105 cupcakes in one day. The head baker predicted he would sell 85 cupcakes that day. What was the percent error of the baker's prediction?

The similarity statement is ΔRSP ≈ ΔRSQ ≈ ΔPRQ.

The proportions are:

QS/QR = QR/PQ

SP/SR = RS/QS

There are ways to prove that two triangles are congruent.

- Side-Side-Side (SSS) Congruence.

The three sides of one triangle are equal to the corresponding three sides of another triangle.

- Side-Angle-Side (SAS) Congruence.

The two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of another triangle.

- Angle-Side-Angle (ASA) Congruence.

The two angles and the included side of one triangle are equal to the corresponding two angles and included side of another triangle.

- Angle-Angle-Side (AAS) Congruence.

The two angles and a non-included side of one triangle are equal to the corresponding two angles and a non-included side of another triangle.

We have,

From the figure,

We have,

Three right triangles:

ΔRSP, ΔRSQ, and ΔPRQ

So,

ΔRSP ≈ ΔRSQ ≈ ΔPRQ

Now,

From ΔRSP and ΔPRQ.

Corresponding sides are considered.

QS/QR = QR/PQ

And,

From ΔRSP and ΔRSQ.

Corresponding sides are considered.

SP/SR = RS/QS

Thus,

The similarity statement is ΔRSP ≈ ΔRSQ ≈ ΔPRQ.

The proportions are:

QS/QR = QR/PQ

SP/SR = RS/QS

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Answer:

The only score that falls outside of this range is 35, so that is the outlier.

Step-by-step explanation:

To find the mean of Allins test scores, we add up all the scores and divide by the number of scores:

Mean = (88+90+35+92+82+90)/6 = 77.83 (rounded to two decimal places)

To find the median, we need to arrange the scores in order from least to greatest:

35, 82, 88, 90, 90, 92

The median is the middle value, which in this case is 89.

To find the mode, we look for the value that appears most frequently. In this case, the mode is 90, since it appears twice.

To find the range, we subtract the lowest score from the highest score:

Range = 92 - 35 = 57

To identify the outlier in the data set, we can use the interquartile range (IQR) and the rule that any data point more than 1.5 times the IQR below the first quartile or above the third quartile is considered an outlier.

First, we need to find the quartiles:

Q1 (first quartile) = 80 (the median of the lower half of the data set)

Q3 (third quartile) = 91 (the median of the upper half of the data set)

IQR = Q3 - Q1 = 11

Any data point more than 1.5 times the IQR below Q1 or above Q3 is considered an outlier.

Lower outlier threshold = Q1 - 1.5(IQR) = 63.5

Upper outlier threshold = Q3 + 1.5(IQR) = 107.5

The only score that falls outside of this range is 35, so that is the outlier.

The value of x in the domain of function f(x) = √(x - 11) will be 13. Then the correct option is A.

A function is an assertion, concept, or principle that establishes an association between two variables. Functions may be found throughout mathematics and are essential for the development of significant links.

The domain means all the possible values of x and the range means all the possible values of y.

The function is given below.

f(x) = √(x - 11)

The value inside the square root should be greater than or equal to zero. Then we have

x - 11 ≥ 0

x ≥ 11

The value of x in the domain of function f(x) = √(x - 11) will be 13. Then the correct option is A.

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Another valid range value for this set of scores is 8, which is the interquartile range.

The range is a measure of the spread or variability of a dataset, and it is calculated as the difference between the largest and smallest values in the set. To find the range for the given set of scores, we first need to order the numbers from smallest to largest:

4, 6, 9, 11, 12, 14, 17

The smallest value in the set is 4, and the largest value is 17, so the range is:

17 - 4 = 13

Therefore, the range for the given set of scores is 13.

The range is a useful measure of variability that is simple to calculate and easy to interpret. However, it has a limitation in that it is sensitive to extreme values or outliers in the dataset. As a result, it may not always provide an accurate representation of the spread of the data.

Another measure of range that works particularly well for variables with precisely defined upper and lower bounds is the interquartile range (IQR). The IQR is calculated as the difference between the 75th percentile (Q3) and the 25th percentile (Q1) of the dataset. It is commonly used with measurements of continuous variables and is less sensitive to extreme values than the range.

To calculate the IQR, we first need to find the median of the dataset. The median is the middle value when the numbers are arranged in order:

4, 6, 9, 11, 12, 14, 17

The median is 11.

Next, we need to find the first and third quartiles. The first quartile (Q1) is the median of the lower half of the dataset, and the third quartile (Q3) is the median of the upper half of the dataset. To find Q1 and Q3, we split the dataset into two halves:

Lower half: 4, 6, 9, 11

Upper half: 12, 14, 17

The median of the lower half is:

(6 + 9) / 2 = 7.5

The median of the upper half is:

(14 + 17) / 2 = 15.5

Therefore, the first quartile (Q1) is 7.5 and the third quartile (Q3) is 15.5.

The interquartile range (IQR) is the difference between Q3 and Q1:

IQR = Q3 - Q1 = 15.5 - 7.5 = 8

Therefore, another valid range value for this set of scores is 8, which is the interquartile range.

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The larger number is 16 and the smaller number is -15.

The numbers are obtained by a system of equations, for which the variables are given as follows:

x and y.

The sum of two numbers is 1, hence:

x + y = 1.

y = 1 - x.

Three times the larger number plus two times the smaller number is 18, hence:

3x + 2y = 18.

Replacing the first equation into the second, the value of x is obtained as follows:

3x + 2(1 - x) = 18

3x + 2 - 2x = 18

x = 16.

Then the value of y is of:

y = 1 - x

y = 1 - 16

y = -15.

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The three-period weighted moving average forecast for period 5 is option (B) 146.4.

In forecasting, the weighted moving average method is commonly used to estimate future values by giving more weight to the recent observations than the older ones.

To calculate the three-period weighted moving average, we need to apply the given weights of 0.5, 0.3, and 0.2 to the last three periods' data. The formula for calculating the weighted moving average is:

Weighted Moving Average = (w₁Y₁) + (w₂Y₂) + (w₃*Y₃)

where w₁, w₂, and w₃ are the weights assigned to the most recent period, second most recent period, and third most recent period, respectively. Y₁, Y₂, and Y₃ are the corresponding values of the periods.

Applying this formula to the given data, we get:

Weighted Moving Average = (0.5148) + (0.3144) + (0.2*146)

= 74 + 43.2 + 29.2

= 146.4

Therefore, the correct option is (B).

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Answer:

Step-by-step explanation:

To find 53 using the equation vw=(2vw)^2 - (v+w)^3, we need to substitute v=5 and w=3 into the equation:

5~3 = (2(5)(3))^2 - (5+3)^3

Simplifying:

5~3 = (30)^2 - (8)^3

5~3 = 900 - 512

5~3 = 388

Therefore, 5~3 is equal to 388.

The set of all points on the real line whose distance from zero is less than 8 can be described by the absolute value inequality:

|x| < 8

where |x| denotes the absolute value of x. This inequality can be read as "the distance between x and 0 is less than 8".

Geometrically, this inequality describes the open interval (-8, 8) on the real number line, which includes all real numbers that are between -8 and 8 but not including -8 and 8 themselves. In other words, the set of all points on the real line that are within 8 units of the origin, excluding the origin itself.

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Type 1 error is rejecting the null hypothesis, even when we know it is true. Here the average diameter is 1.65mm, but we decide that it is not 1.65mm.

The data which given says that the average diameter should be 1.65mm. Bigger and smaller diameters may result in problems like pain and rupture of RBCs.

So the null and alternate hypothesis are

Null hypothesis , H₀ : μ= 1.65 mm

Alternate hypothesis, H₁ : μ≠ 1.65 mm

Here the sample size, n= 35, σ = 0.07, mean = 1.64

Type 1 error means rejecting null hypothesis, when it is true. It is also known as false positive error. A false positive can occur if something other than the stimuli applied during the test is resulting the outcome. This cannot be avoided because of the degree of uncertainty involved.

In type 1 error, we assume that the null hypothesis, H₀ : μ=1.65 is false, that means the average diameter is not 1.65 mm, even if the actual diameter is 1.65 mm.

Se decide average diameter is not 1.65 mm.

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The simplification of the given expression

[tex]5 \sqrt[3]{250} - 3 \sqrt[3]{686} [/tex]

is

[tex] = 4 \sqrt{2} [/tex]

[tex]5 \sqrt[3]{250} - 3 \sqrt[3]{686} [/tex]

[tex] = 5 \sqrt[3]{125 \times 2} - 3 \sqrt[3]{343 \times 2} [/tex]

[tex] = 5 \times 5 \sqrt{2} - 3 \times 7 \sqrt{2} [/tex]

[tex] = 25 \sqrt{2} - 21 \sqrt{2} [/tex]

[tex] = 4 \sqrt{2} [/tex]

Consequently,

[tex]4 \sqrt{2} [/tex]

is the answer to the given expression.

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Given that the person is colorblind, the likelihood that they are male is around 0.926, or 92.6%.

As per the question given,  

Let's use Bayes' theorem to solve this problem.

Let M be the event that the person is male, and C be the event that the person is color-blind. We want to find the probability P(M|C), which is the probability that the person is male given that they are color-blind.

Using Bayes' theorem, we have:

P(M|C) = P(C|M) * P(M) / P(C)

We know that P(M) = 0.5, since 50% of the population are men. We also know that P(C|M) = 0.05, since 5% of men are color-blind. To find P(C), we can use the law of total probability:

P(C) = P(C|M) * P(M) + P(C|F) * P(F)

where F is the event that the person is female. We know that P(F) = 0.5, since 50% of the population are women, and we also know that P(C|F) = 0.004, since 0.4% of women are color-blind.

Substituting the values into Bayes' theorem, we have:

P(M|C) = 0.05 * 0.5 / (0.05 * 0.5 + 0.004 * 0.5)

= 0.926

Therefore, the probability that the person is male given that they are color-blind is approximately 0.926, or 92.6%.

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As a result, V is a vector space over the real number field.

As per the question given,

Yes, V is a vector space over the field of real numbers with the given operations.

To show this, we need to verify that the following vector space axioms are satisfied for any vectors u, v, and w in V and any scalars a and b in F:

All of these axioms are easily verified using the given operations. Therefore, V is a vector space over the field of real numbers.

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Answer:

  x = 1

Step-by-step explanation:

You want the solution to the equation √(x²+2x+6) +x² = √(2x+2) -x +3.

We find a graphing calculator to be a useful tool for finding the solutions to equations like this. It shows the solution to be x = 1.

Solved in the usual way, this would resolve to an 8th degree equation with one integer factor: (x -1). The remaining 7th degree polynomial factor has 3 real roots and 4 complex roots. All of these 7 irrational solutions are extraneous. (They do not satisfy the original equation.)

The solution is x = 1.

__

Additional comment

Our approach to removing the radicals is to ...

The usual approach to looking for roots is to use Descartes' rule of signs and the Rational Root theorem.

You can determine there are 3 or 1 positive real roots, and 5, 3, or 1 negative real roots. The only rational real roots will be ±1 or ±47, and we know that ±47 can't work. (47^8 will not be balanced by any of the other terms in the polynomial.)

You can see already that the graphing calculator approach is much preferred.

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The unit fraction is given by A = 2/2 and the whole number is represented by B = 4/2 = 2

An element of a whole is a fraction. The number is represented mathematically as a quotient, where the numerator and denominator are split. Both are integers in a simple fraction. A fraction appears in the numerator or denominator of a complex fraction. The numerator of a proper fraction is less than the denominator.

Given data ,

Let the first fraction be represented as A

Now , the unit fraction is a fraction which has the value of 1

And , the numbers are represented by set S = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 }

The unit fraction A = 1/1 = 2/2 = 3/3

Let the whole fraction be represented as B

Now , the value of B is

B = 4/2 = 2

B = 6/3 = 2

Therefore , the value of A and B are 1/1 and 4/2 respectively

Hence , the fractions are solved

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He had 48.75 oz of salsa.

In mathematics, multiplication is a method of finding the product of two or more numbers. It is one of the basic arithmetic operations, that we use in everyday life.

here, we have,

I guess that Clay is filling 6.5 oz bottles of salsa.

And we know that he was able to fill 7.5 of these bottles,

then if each bottle has 6.5 oz, 7.5 of them have a total mass of salsa will be equal to 7.5 times 6.5 ounces,

this is:

M = 7.5*6.5 oz = 48.75 oz

This means that Clay had 48.75 ounces of Salsa

A drawn model to show this is seen below.

Each big square represents 0.5 oz,

and the area black delimited area represents a total amount of 6.5 oz, we have 7 of them, then you can draw that 7 times,

and count the squares,

and then you will find the total ounces of salsa that Clay had, which is equivalent to the multiplication we did above.

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Answer:

Step-by-step explanation:

If you are using both y & x they are easy to find

distance from 9 and 2 is seven units and the distance from each other

3 & 2 are one unit hoped it helped you so you can find the miles

When a research is conducted on a randomly chosen representative sample, hypothesis testing involves gathering data and calculating how likely a specific set of data is. No connection exists between the factors.

The t test is the most used null test for this statistical connection. Only one t test, the reliant t test, as well as the individual t test are three different t test types that are examined in this section and are used to a variety of study methods.

The probability that you would have discovered a certain collection of data if the null hypothesis had been correct is expressed as a number called the p value, which is determined from a statistical test. In order to determine either to reject a null hypothesis, P values are utilized in hypothesis testing.

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The correct option regarding the number of x-intercepts of each function is given as follows:

Both functions have the same number of x-intercepts.

The x-intercept of a function is given by the value of x when f(x) = 0, that is, the value of x when the function crosses the x-axis.

From the Table, for function A, these values are given as follows:

x = 1 and x = 3.

For Function B, the intercepts are given as follows:

|x| - 2 = 0

|x| = 2

x = -2 or x = 2.

The intercepts are different, but both functions have the same number of x-intercepts.

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Answer:

Both functions have the same number of x-intercepts.

Step-by-step explanation: Did the test, got it correct.

If f(x), g(x), and h(x) are functions such that f(x) is θ(g(x)) and g(x) is θ(h(x)), it is shown that f(x) is θ(h(x)).

Since f(x) is θ(g(x)), there exist constants A, B > 0 such that

A × g(x) < f(x) < B × g(x) for sufficiently large x.

Similarly, since g(x) is θ(h(x)), there exist constants C, D > 0 such that

C × h(x) < g(x) < D × h(x) for sufficiently large x.

Hence, for a sufficiently large x, we have

f(x) < B g(x) < B × (D × h(x)) = (BD) h(x), and

f(x) > A g(x) > A × (C × h(x)) = (AC) h(x).

Hence, we have constants E, F > 0 (where E = AC and F = BD) such that

E × h(x) < f(x) < F × h(x) for sufficiently large x.

Therefore, f(x) is θ(h(x)).

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The volume of the rectangular tank is 4 feet * 5 feet * 4 feet = 80 cubic feet.

(a) To empty half of the tank, we need to pump out 80/2 = 40 cubic feet of water.

The weight of the water being pumped out is 40 cubic feet * 62.4 pounds/cubic foot = 2528 pounds.

To pump the water out over the top edge, work has to be done against gravity, which means the work done is equal to the weight of the water times the height it's lifted, which is 4 feet.

Therefore, the work done in pumping half of the water out is 2528 pounds * 4 feet = 10112 foot-pounds.

(b) To empty the tank, we need to pump out 80 cubic feet of water.

The weight of the water being pumped out is 80 cubic feet * 62.4 pounds/cubic foot = 5056 pounds.

To pump the water out over the top edge, work has to be done against gravity, which means the work done is equal to the weight of the water times the height it's lifted, which is 4 feet.

Therefore, the work done in pumping all of the water out is 5056 pounds * 4 feet = 20224 foot-pounds.

The complete question is:-

A rectangular tank with a base of 4 feet by 5 feet and a height of 4 feet is full of water. The water weighs 62.4 pounds per cubic foot. How much work is done in pumping water out over the top edge in order to empty (a) half of the tank and (b) all of the tank?

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The rounded numbers for the given decimal numbers are 4, 8 and 9.

To change a number into an approximation having fewer significant digits, is called the rounding numbers.

For example :- Round off 15.4 to 15, round off 15.51 to 15.5 or to 16, round off 0.499 to 0, and round off 970,000 to 1 million.

Given are three decimal numbers, 3.896, 7.8143 and 9.3959 we are asked to round them to whole number.

The rule for rounding the numbers is :-

Example: 38 rounded to the nearest ten is 40.

Example: 33 rounded to the nearest ten is 30.

Therefore, in 3.896, the number after the decimal is 8 which is greater than 5 so rounding up we will get, 4

Similarly, 7.8143 = 8

And in 9.3959, the number after the decimal is 3 which is smaller than 5 so rounding down we will get, 9

Hence, the rounded numbers for the given decimal numbers are 4, 8 and 9.

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The figure repeats in the tessellation is Figure B, the correct option is A

A regular tessellation is a pattern made by repeating a regular polygon. In simpler words regular tessellations are made up entirely of congruent regular polygons all meeting vertex to vertex.

How many regular tessellation are possible?

There are only 3 regular tessellation.

1. Triangle

2. Square

3. Hexagon

Given;

The figure

Not more than 3 regular tessellations are possible because the sums of the interior angles are either greater than or less than 360 degrees;

Therefore, repeated tessellation will be B.

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Answer:

16. The rational expression 3/(x-5) is undefined when x equals 5.

17.  The rational expression 8/(x+2) is undefined when x equals -2.

Step-by-step explanation:

16.

The rational expression 3/(x-5) is undefined when the denominator (x-5) is equal to zero, because division by zero is undefined. Therefore, we need to find the values of x that make the denominator zero.

To find those values, we can set the denominator equal to zero and solve for x:

x - 5 = 0

x = 5

Therefore, the rational expression 3/(x-5) is undefined when x equals 5.

17.

The rational expression 8/(x+2) is undefined when the denominator (x+2) is equal to zero, because division by zero is undefined. Therefore, we need to find the values of x that make the denominator zero.

To find those values, we can set the denominator equal to zero and solve for x:

x + 2 = 0

x = -2

Therefore, the rational expression 8/(x+2) is undefined when x equals -2.

The Population parameter is population mean i.e μ. The Interpretation of 90% confidence that population mean i.e μ lies between 116.8 and 149.2.

We have, a Normalised average amount of time, in minutes, data for students to complete a standardized test.

Sample size, n = 49

Significance level, = 90% = 0.90

Confidence interval is range of values. It is calculated by below formula, confidence interval,

CI = Mean ± Z( standard deviations/sample size)

but we have, the 90% confidence interval is equals to [116.8,149.2]. The population is a large set and it is deficult to work on population, so sample subsets of it are considered for observation. Population parameter is population mean. The interpretation from provide informations about 90% of confidence interval is that population mean lie between 116.8 and 149.2, i.e., 116.8 ≤ μ ≤149.2.

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Adam makes his first mistake in the first step. And the product of linear factors will be (√5x - 2i √2) and (√5x + 2i√2).

It is a method for dividing a polynomial into pieces that will be multiplied together. At this moment, the polynomial's value will be zero.

The expression is given below.

⇒ 4 + 5x² + 4

Simplify the expression, then we have

⇒ 5x² + 8

Factorize the expression, then we have

⇒ 5x² + 8

⇒ (x√5)² - (2i √2)²

⇒ (√5x - 2i √2) (√5x + 2i√2)

Adam makes his first mistake in the first step. And the product of linear factors will be (√5x - 2i √2) and (√5x + 2i√2).

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The expected demand for the high-pressure cleaner is 1.86.

Frequency distribution is a statistical tool that organizes data into different categories and shows how often each category occurs. In this case, the demand for a high-pressure cleaner is the category, and the number of times it occurs is the frequency.

To calculate the expected demand, we need to first find the relative frequency of each demand level. The relative frequency is the percentage of the total number of demands accounted for by each demand level. We can calculate the relative frequency by dividing the frequency of each demand level by the total number of demands.

Total number of demands = 20 + 23 + 27 + 14 + 16 = 100

Relative frequency of demand level 0 = 20/100 = 0.20

Relative frequency of demand level 1 = 23/100 = 0.23

Relative frequency of demand level 2 = 27/100 = 0.27

Relative frequency of demand level 3 = 14/100 = 0.14

Relative frequency of demand level 4 = 16/100 = 0.16

Next, we multiply each demand level by its relative frequency and add the products to get the expected demand.

Expected demand = (0 × 0.20) + (1 × 0.23) + (2 × 0.27) + (3 × 0.14) + (4 × 0.16)

=> 1.86

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Complete Question:

The following data has been recorded regarding the rental history of a high pressure cleaner. Find the expected demand for the high pressure cleaner.

Demand Frequency

0              20

1               23

2              27

3              14

4              16

Answer:

Step-by-step explanation:

since y = -8x + 6

if x=-10, y=-8x10+6=-74, so A is not the solution

if x =-8, y=-8x(-8)+6=70, so B is not the solution

if x =5, y=-8x5+6=-34, so C is the solution

if x =3, y=-8x3+6=-18, so D is not the solution

The probability of failure on the trial 9 for the given binomial experiment is 0.15.

An experiment with a set number of independent trials and just two results is referred to as a binomial experiment. The results of these experiments can be classified as either successes or failures. Due to the nature of the experiment's subject, results may only be categorised as successes or failures.

Due to the set number of outcomes that might occur in each experiment's trial, binomial experiments are unique from other types of studies. Binomial experiments, in particular, can only ever produce one of two outcomes.

The probability of success and failure of a trial is same for every number of trials.

Given that,

n = 12 trials

Success = 0.85

The failure is given as:

failure = 1 - success of trial

failure = 1 - 0.85

failure = 0.15

Hence, the probability of failure on the trial 9 for the given binomial experiment is 0.15.

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