Mark had 5/6 as many birds stickers as bear stickers at first. After buying another 20 stickers of each type, he has 7/8 as many bird stickers as bear stickers. How many of each type of stickers did mark have at first?

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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After 1020 sec Kristen get her Dinner ready.

The same attribute is expressed using a unit conversion, but in a different unit of measurement.

For instance, time can be expressed in minutes rather than hours, and distance can be expressed in kilometres, feet, or any other measurement unit instead of miles.

Subtract Kristen's arrival time from the time her food is ready to get the number of seconds she needs to wait before eating.

So,

5 :51 pm - 5: 34 pm = 17 min

As, 1 minute = 60 minute

So, 17 minutes

= 17(60)

= 1020 sec

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The expression for the perimeter of the triangle 6x+17.

Perimeter the sum of length of the sides used to made the given figure. A regular figure with n-sides has n equal sides in it, and they are the only parts of it(that means, nothing more than those equal lengthened n sides).

We have Given the length of the bottom of the triangle = x.

Since other side is 11 more than four times the length of the bottom of the triangle.

Also we have last side is 6 more than the bottom of the triangle.

Now, the sides of triangle;

x, 4x+11, x+6

Perimeter is the sum of lengths of all sides of figure.

x+ 4x+11+ x+6

=6x+17

Therefore, the perimeter expression will be 6x+17.

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The piecewise function is continuous only if a = 2.

The function will be continuous if both parts of the function have the same value on the "jump" between the two parts.

The first part evaluated in x = 1 is:

f(x) = (4x^3 - 4x)/(x - 1)

We can rewrite the numerator as:

(4x^3 - 4x) = 4x*(x^2 - 1) = 4x*(x + 1)*(x - 1)

Then we can rewrite the function as:

f(x) = 4x*(x + 1)

Evaluating in x = 1 we will get:

f(1) = 4*1*(1 + 1) = 8

Then the other piece evaluated in x = 1 also must be equal to 8, we will get:

f(x) = 4x^2 + 2x + a

f(1) = 4*1^2 + 2*1 + a = 8

    = 4 + 2 + a  = 8

       a = 8 - 4 - 2 = 2

The value of a must be 2.

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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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The area of sail is equivalent to (4x + 8x + 3) square units.

The area of a triangle is given as -

[tex]$A_{T} =\frac{1}{2} \times b \times h[/tex]

Given is a sail whose base is (2x + 1) units and height is (4x + 6) units.

We can write the area of sail as -

A = 1/2 x b x h

A = 1/2 x (2x + 1) x (4x + 6)

A = (2x + 1)(2x + 3)

A = 2x(2x + 3) + 1(2x + 3)

A = 3 + 2x + 6x + 4x²  

A = 4x² + 8x + 3

Therefore, the area of sail is equivalent to (4x² + 8x + 3) square units.

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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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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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(a) 0.5, (b) f_Y(y) = 12y(1-y), (c) 9/128, (d) 7/36 - Probability calculations using joint density function.

(a) To find the Probability that the Turkish tobacco represents over a portion of the mix, we want to incorporate the joint thickness capability over the district where x > 0.5 and y < 1 - x. This gives:

P(X > 0.5) = ∫[0.5,1]∫[0,1-x] 24xy dy dx = 0.5

Consequently, the Probability that the Turkish tobacco represents over around 50% of the mix is 0.5.

(b) To find the negligible thickness capability for the extent of homegrown tobacco, we really want to incorporate the joint thickness capability over all potential upsides of Turkish tobacco:

f_Y(y) = ∫[0,1-y] 24xy dx = 12y(1-y) ; 0 < y < 1

Thusly, the peripheral thickness capability for the extent of homegrown tobacco is f_Y(y) = 12y(1-y).

(c) To find the Probability that the extent of Turkish tobacco is under 1/8 given that the mix contains 3/4 homegrown tobacco, we really want to utilize Bayes' standard:

P(X < 1/8 | Y = 3/4) = P(X < 1/8 ∩ Y = 3/4)/P(Y = 3/4)

We can find the numerator by coordinating the joint thickness capability over the district where x < 1/8 and y = 3/4:

∫[0,1/8] 24xy dx = 27/128

To find the denominator, we coordinate the joint thickness capability over all potential upsides of Turkish tobacco when y = 3/4:

∫[0,1/4] 24xy dx = 3

Thusly, the contingent Probability is:

P(X < 1/8 | Y = 3/4) = (27/128)/3 = 9/128

(d) To find the likelihood that the extent of homegrown tobacco is over two times the extent of the Turkish tobacco, we want to incorporate the joint thickness capability over the area where y > 2x and x + y < 1. This gives:

P(Y > 2X) = ∫[0,1/3]∫[2x,1-x] 24xy dy dx + ∫[1/3,1/2]∫[2x,1-x] 24xy dy dx

= 7/36

Accordingly, the likelihood that the extent of homegrown tobacco is over two times the extent of the Turkish tobacco is 7/36

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The complete question is:

A tobacco company produces blends of tobacco, with each blend containing various proportions of Turkish, domestic, and other tobaccos. The proportions of Turkish and domestic in a blend are random variables with joint density function (X = Turkish and Y = domestic) 24xy ;0< x, y < 1, x +y<1 fx.y(1,7) (2) ; otherwise (a) (1 point) Find the probability that in a given box the Turkish tobacco accounts for over half the blend. Page 2 (b) (1 point) Find the marginal density function for the proportion of the domestic tobacco. (c) (1 point) Find the probability that the proportion of Turkish tobacco is less than 1/8 if it is known that the blend contains 3/4 domestic tobacco. (d) (2 points) Find the probability that the proportion of domestic tobacco is more than twice the proportion of the Turkish tobacco.

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

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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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 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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31 in by 31/4 in is the required dimension of the poster

The formula for calculating the area of a rectangle is expressed as:

A = lw

If the length of a rectangular poster is 4 more inches than three times its width, then;

l = 3w - 4

Substitute

340 = w(3w - 4)

340 = 3w^2 - 4w

3w^2 - 4w - 340 =0

On factorizing, the measure of the width of the poster is 34/3 in

l = 3w - 4

l = 3(34/3) - 4
l = 31in

Hence the dimension of the given poster is 31 in by 31/4 in

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The solution to the equation x - 14 = 22 is x = 36

An equation is an expression that shows the relationship between numbers and variables using mathematical operators like addition, subtraction, multiplication and division. Equations can be linear, quadratic.

Let x represent the number. 14 less than a number is 22, hence:

x - 14 = 22

Add 14 to both sides of the equation:

x - 14 + 14 = 22 + 14

x = 36

The solution to the equation is x = 36

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A programmer erroneously wrote an expression as 0 < x < 10. This expression 0 < x < 10 can be rewritten using logical AND as (0 < x) AND (x < 10).

The expression 0 < x < 10 is commonly used to indicate that x is a value between 0 and 10, but it is not correct in programming. In programming, logical operations like "less than" and "greater than" are usually represented by < and > symbols, respectively.

To properly express the condition that x is between 0 and 10, we need to use the logical operator AND, which allows us to combine multiple conditions.

The logical AND operator in programming is typically represented by the symbol && or the word "AND". It evaluates to true if and only if both of the conditions on either side of the operator are true. If either condition is false, the whole expression is false.

To rewrite the expression 0 < x < 10 using logical AND, we first separate the conditions into two separate expressions:

x > 0 AND x < 10

which can be written as,

(0 < x) AND (x < 10).

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Answer:

there are 462,462 different ways the committee can be formed.

Step-by-step explanation:

To form the committee, we need to choose 5 teachers out of 11 and 4 students out of 14. We can use the combination formula to find the number of ways:

Number of ways to choose 5 teachers out of 11:

C(11, 5) = 11! / (5! * 6!) = 462

Number of ways to choose 4 students out of 14:

C(14, 4) = 14! / (4! * 10!) = 1001

The total number of ways to form the committee is the product of these two combinations:

462 * 1001 = 462462

Therefore, there are 462,462 different ways the committee can be formed.

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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A. The function to represent the balance in the account after t years is:

[tex]A = 1500(1 + (0.012 / 4))^{(4t)}[/tex]

B. The balance in the account after 3 years will be $1566.66.

C. The balance in the account after 6 years will be $1639.98.

Compound interest is defined as interest paid on the original principal and the interest earned on the interest of the principal.

A = P(1+r/100)ⁿ

A. The function to represent the balance in the account after t years can be represented as:

[tex]A = 1500(1 + (0.012 / 4))^{(4t)}[/tex]

where A is the balance in the account after t years, 1500 is the initial deposit, 0.012 is the annual interest rate as a decimal, 4 is the number of compounding periods per year, and (0.012 / 4) is the interest rate per compounding period.

B. To find the balance after 3 years:

A = 1500(1 + (0.012 / 4))⁴ˣ³ = 1500 (1 + (0.012 / 4))¹²

A = 1500 × 1.0444

A = 1566.66

So, the balance in the account after 3 years will be $1566.66.

C. To find the balance after 6 years:

A = 1500 (1 + (0.012 / 4))⁴ˣ⁶ = 1500 * (1 + (0.012 / 4))²⁴

A = 1500 × 1.0932

A = 1639.98

So, the balance in the account after 6 years will be $1639.98.

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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:

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

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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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 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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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 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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