WILL GIVE BRAINLIEST
Given the parent function g(x) = log2 x
What is the equation of the function shown
in the graph?

The Equation of the line which is parallel to the line "y = 3x + 6", and also passes through (4,7) is "y = 3x - 5".

In order to find an equation of a line which passes through point (4,7) and is parallel to line "y = 3x + 6", we use the fact that parallel lines have the same slope.

The given line ""y = 3x + 6" has a slope of 3,

So, the "parallel-line" we want to find must also have a slope of 3.

Now, by using the "point-slope" form of the equation of a line, which is : y - y₁ = m(x - x₁),

where m is slope and (x₁, y₁) is a point on the line,

So, we substitute "m = 3" and (x₁, y₁) = (4,7) to get:

⇒ y - 7 = 3(x - 4),

⇒ y - 7 = 3x - 12,

⇒ y = 3x - 5

Therefore, the equation of the required line is y = 3x - 5.

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Since the population standard deviation is unknown, we use a t-distribution to find the critical value. The degrees of freedom for the t-distribution is n-1, where n is the sample size. In this case, n = 7, so the degrees of freedom is 7-1 = 6. The critical value for a t-distribution with 6 degrees of freedom and a significance level of α = 0.200 (two-tailed) can be found using a t-table or calculator. The critical value is approximately ±1.94.

part a.

the percentage of eggs between 42 and 45mm is 48.48%

part b.

The median width is approximately (42+45)/2 = 43.5mm.

The median length is approximately (56+59)/2 = 57.5mm.

part c.

The width of grade A chicken eggs has a range of about 24mm.

part d.

I think its impossible to determine because we don't have the value for the standard deviation.

The second option should be correct.

A histogram is described as an approximate representation of the distribution of numerical data.

part a.

From the histogram, we see that the frequency for the bin that ranges from 42 to 45mm is 4 and we have  a total of 33 eggwe use this values and calculate the percentage of eggs between 42 and 45mm is 48.48%.

part b.

we have an  estimation  that the median of the width is 48mm and the median of the length is around 60mm.

part c.

Also from the histogram, we notice  that the smallest value is around 36mm and the largest value is around 66mm, hence the width of grade A chicken eggs has a range of about 24mm.

In a histogram, the range is the width that the bars cover along the x-axis and these are approximate values because histograms display bin values rather than raw data values.

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The correct arrangement of problem is explained below and their solution are as follows:

(1) Zaira will arrive at her grandmother's house at 9:00 am.

(2) The total distance covered by bus is 207.5 km.

(3) The average-speed of the train was 28 km/h.

Part (1) : The Problem is : Zaira goes to her grandmother's house. If she leaves home at 6:00 in the morning, she cycles 30 km at a steady speed of 10 km. What time will she arrive?

Solution:

Zaira cycles at a steady speed of 10 km, she will cover the distance of 30 km in 30/10 = 3 hours.

So, she will arrive at her grandmother's house at 6:00 + 3:00 = 9:00 am.

Part (2) : Problem : A bus had an average speed of 65 kph for 1.5 hours in the morning. It had average speed of 55 kph for 2 hours in afternoon. What was total distance covered by bus?

Solution:

The distance covered by the bus in the morning can be calculated as:

Distance = Speed × Time = 65 kph × 1.5 hours = 97.5 km,

The distance covered in the afternoon can be calculated as:

Distance = Speed × Time = 55 kph × 2 hours = 110 km

So, total-distance covered by bus is = 97.5 km + 110 km = 207.5 km.

Part (3) : Problem : A train left Station X at 9:00 a.m. and arrived station Y at 9:30 a.m. The distance between the two stations is 14 km. What was  average speed of train?

Solution:

The time taken by the train to cover the distance of 14 km can be calculated as:

Time = Arrival Time - Departure Time = 9:30 am - 9:00 am = 0.5 hours

The average speed of the train = Distance/Time = 14 km/0.5 hours = 28 km/h;

Therefore, the average speed of the train was 28 km/h.

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The given question is incomplete, the complete question is

Directions: Arrange the sentences in the box to form a problem. Then solve each problem.

(1) If she leaves home at 6:00 in the morning

What time will she arrive?

Zaira goes to her grandmother's house

She cycles 30 km at a steady speed of 10 km

(2) I had an average speed of 55 kph for 2 hours in the afternoon

What was the total distance covered by the bus

A bus had an average speed of 65 kph for 1. 5 hours in the morning.

(3) What was the average speed of the train?

The distance between the two stations is 14 km

A train left Station X at 9:00 a.m. and arrived station Y at 9:30 a.m. ​

The graph of the exponential function [tex]y = 2^{3x}[/tex] is given as follows:

An exponential function has the definition presented as follows:

[tex]y = ab^x[/tex]

In which the parameters are given as follows:

The function in this problem is given as follows:

[tex]y = 2^{3x}[/tex]

Hence the parameter values are:

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Answer: 5

2.75+(2.45x5) = 15

Answer:

If Mohal waits on 16 tables, he can expect to earn $25 wages + ($15 tips x 16 tables) = $265 in a day.

If Mohal waits on 1 table, he can expect to earn $25 wages + ($15 tips x 1 table) = $40 in a day.

Answer:no image

Step-by-step explanation:

Answer:

Step-by-step explanation:

tried my best as the first two got me stuck

wouldn't recommend trying my answer and Id wait for another answer

A figure with slope of consecutive sides -2/3 and 3/2 is a rectangle and it is not a parallelogram.

To prove that it is a parallelogram or a rectangle, we need to show that the opposite sides are parallel and the adjacent sides are perpendicular.

Let's first check if the opposite sides are parallel. The slope of one side is -2/3, and the slope of the adjacent side is 3/2. For opposite sides to be parallel, the slopes must be equal. However, -2/3 and 3/2 are not equal, so we can conclude that the given figure is not a parallelogram.

Now, let's check if the adjacent sides are perpendicular. The product of the slopes of the adjacent sides is

(-2/3) x (3/2) = -1, which is the slope of a line perpendicular to both sides. Since the product of the slopes is -1, we can conclude that the adjacent sides are perpendicular.

Therefore, figure is not a parallelogram, but it is a rectangle.

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Check the picture below.

[tex]\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2 \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{6.6+x}\\ a=\stackrel{adjacent}{6.6}\\ o=\stackrel{opposite}{8.8} \end{cases} \\\\\\ (6.6+x)^2= (6.6)^2 + (8.8)^2\implies (6.6+x)^2=121\implies (6.6+x)^2=11^2 \\\\\\ 6.6+x=11\implies x=4.4[/tex]

The measures of the arc angle AP is 63° using the Angles of Intersecting Chords Theorem

The Angles of Intersecting Chords Theorem states that the angle formed by the intersection of the chords is equal to half the sum of the intercepted arcs, and conversely, that the measure of an intercepted arc is half the sum of the two angles that intercept it.

109° = (AP + RQ)/2

109 = (AP + 155)/2

AP + 155 = 2 × 109 {cross multiplication}

AP + 155 = 218

AP = 218 - 155 {collect like terms}

AP = 63°

Therefore, the measures of the arc angle AP is 63° using the Angles of Intersecting Chords Theorem

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The probability that a given voicemail is between 20 and 50 seconds is 0.8185, or 81.85%.

To find the probability that a voicemail is between 20 and 50 seconds, we need to standardize the values and use a standard normal distribution table.

First, we find the z-scores for 20 seconds and 50 seconds:

z1 = (20 - 40) / 10 = -2

z2 = (50 - 40) / 10 = 1

Using a standard normal distribution table, we can find the area to the left of each z-score:

Area to the left of z1 = 0.0228

Area to the left of z2 = 0.8413

To find the probability between 20 and 50 seconds, we subtract the area to the left of z1 from the area to the left of z2:

P(20 < x < 50) = P(-2 < z < 1)

= 0.8413 - 0.0228

= 0.8185

Therefore, the probability that a given voicemail is between 20 and 50 seconds is 0.8185, or 81.85%.

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To model the volume of a rectangular prism with length 26cm and width w and height h such that the sum of the width and height is 32cm, we can use the following function:

V(w, h) = 26wh

subject to the constraint:

w + h = 32

We can solve for one of the variables in the constraint equation and substitute it into the volume equation, giving us:

w + h = 32  =>  h = 32 - w

V(w) = 26w(32 - w) = 832w - 26w^2

To find the maximum possible volume, we can take the derivative of this function with respect to w and set it equal to zero

dv/dw= 832 - 52w = 0

Solving for w, we get:

w = 16

Substituting this value back into the constraint equation, we get:

h = 32 - w = 16

Therefore, the maximum possible volume of the prism is:

V(16, 16) = 26(16)(16) = 6656 cubic cm

So the function to model the volume of the rectangular prism is V(w) = 832w - 26w^2, and the maximum possible volume is 6656 cubic cm when the width and height are both 16cm.

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The area of the shaded sector of the given circle would be = 42,593.5 in²

To calculate the area of the given sector the formula that should be used is given as follows;

The area of a sector =( ∅/2π) × πr²

where;

π = 3.14

r = 12 in

∅ = 60°

Area of the sector = (60/2×3.14)b × 3.14× 12×12

= 94.2× 452.16

= 42,593.5 in²

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Therefore, the missing value of G is 7.608 cm, which is option A.

If G is two and one-half times smaller than 19.02 cm, we can find the value of G by multiplying 19.02 cm by 2/5, since two and one-half is equal to five halves, or 2/5 when expressed as a fraction.

G = (2/5) x 19.02 cm

Simplifying this expression:

G = 7.608 cm

Therefore, the missing value of G is 7.608 cm, which is option A.

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Given two events E and F with Pr(E) = 0. 4, Pr(F) = 0. 5, and Pr(EnF) = 0. 3 then:

a) Pr(E|F) = 0.6

b) Pr(F|E) = 0.75

c) NO, E and F are not independent.

a) To find Pr(E|F), we use the formula: Pr(E|F) = Pr(EnF)/Pr(F). Substituting the given values, we get Pr(E|F) = 0.3/0.5 = 0.6.

b) Similarly, to find Pr(F|E), we use the formula: Pr(F|E) = Pr(EnF)/Pr(E). Substituting the given values, we get Pr(F|E) = 0.3/0.4 = 0.75.

c) We can check for independence by seeing if Pr(E) = Pr(E|F) or Pr(F) = Pr(F|E). However, since Pr(E) ≠ Pr(E|F) and Pr(F) ≠ Pr(F|E), we can conclude that E and F are not independent.

In other words, the occurrence of one event affects the probability of the other event occurring. Specifically, the fact that Pr(EnF) ≠ Pr(E)Pr(F) indicates that the events are dependent.

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An expression for the sequence of operations described "Subtract three from the product of seven and eight." is (7 × 8) - 3.

In order to evaluate and solve this expression, we would have to apply the PEMDAS rule, where mathematical operations within the parenthesis (grouping symbols) are first of all evaluated, followed by exponent, and then multiplication or division from the left side of the equation to the right. Lastly, the mathematical operations of addition or subtraction would be performed from left to right.

Based on the information provided, we have the following mathematical expression:

Expression: (7 × 8) - 3

56 - 3

53

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The main threat to concluding that the tutor was effective in helping students understand confounding and obscuring variables is the issue of 'practice effects'.

Practice effects occur when exposure to specific materials or questions during a study can artificially inflate performance on subsequent measures, such as an exam.

In this case, if the practice questions from the tutor were exactly the same as those on the exam.

Then the students who interacted with the tutor may have had an advantage on the exam .

Due to familiarity with the questions rather than a true understanding of the material.

The use of the same practice questions in the tutor and on the exam .

It represents a potential confounding variable that could affect the interpretation of the results.

It is possible that the tutor may have been effective.

But it is also possible that the practice effect of the repeated questions contributed to the observed improvement in performance.

Future research could use different practice questions in the tutor and on the exam.

Use a different exam altogether to assess performance.

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Answer:

It will cost $288 to carpet the living room at $36 per square yard.

Step-by-step explanation:

First, we need to convert the room dimensions to square yards, since the carpet price is given in square yards.

The area of the living room is:

[tex]\sf:\implies 6\: ft \times 12\: ft = 72\: ft^2[/tex]

To convert this to square yards, we divide by 9 (since there are 9 square feet in a square yard):

[tex]\sf:\implies \dfrac{72\: ft^2}{9} = 8\: yards^2[/tex]

So the living room is 8 square yards in area.

To find the cost of carpeting the room, we multiply the area by the cost per square yard:

[tex]\sf:\implies 8\: yards^2 \times \$36/square\: yard = \boxed{\bold{\:\:\$288\:\:}}\:\:\:\green{\checkmark}[/tex]

Therefore, it will cost $288 to carpet the living room at $36 per square yard.

Using the given function, we can plug in x=3 and t=5 to find the concentration of the drug in the blood:

C = f(x,t) = 26te^-(5-x)
C = f(3,5) = 26(5)e^-(5-3)
C = f(3,5) = 130e^-2

Using a calculator, we can simplify this to:

C = f(3,5) ≈ 32.22 mg/liter

Therefore, the concentration of the drug in the blood 3 hours after injection with a dosage of 5 mg is approximately 32.22 mg/liter.
Hi! To find the concentration C at f(3,5), you'll need to plug in the values for x and t into the given function f(x,t) = 26te^-(5-x).

So, f(3,5) = 26(5)e^-(5-3) = 130e^(-2).

Now, calculate the exponential value: e^(-2) ≈ 0.1353.

Finally, multiply this value by 130: 130 * 0.1353 ≈ 17.589.

Thus, f(3,5) = 17.589 mg/liter (rounded to 3 decimal places).

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The explicit equation for the nth term of the arithmetic sequence is an = 9 - 2.7n.

What is the implicit equation?

An implicit equation is an equation in which the variables are not explicitly expressed in terms of each other. In other words, the equation does not give a direct formula for one of the variables in terms of the other(s), but rather relates the variables through some function or equation.

What is the explicit equation?

An explicit equation is an equation in which one variable is expressed directly in terms of the other(s). In other words, the equation gives a formula for one of the variables in terms of the other(s).

According to the given information:

the first term of the sequence is a1 = 6.3, and the common difference between consecutive terms is d = -2.7 (since we subtract 2.7 from each term to get to the next term). Therefore, the explicit equation for the nth term of the sequence is:

an = 6.3 + (n - 1)(-2.7)

Simplifying this expression, we get:

an = 6.3 - 2.7n + 2.7

an = 9 - 2.7n

So the correct equation for the nth term of the arithmetic sequence 6.3, 3.6, 0.9, -1.8, -4.5, ... is:

an = 9 - 2.7n

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The probability that there will be at least 4 typos on page 301 is 0.847

To solve this problem, we need to make some assumptions. Let's assume that the number of typos on each page follows a Poisson distribution with a mean of 6 typos per page, and that the number of typos on one page is independent of the number of typos on any other page.

Under these assumptions, we can use the Poisson distribution to calculate the probability of observing a certain number of typos on a given page.

Let X be the number of typos on page 301. Then X follows a Poisson distribution with a mean of 6 typos per page. The probability of observing at least 4 typos on page 301 can be calculated as follows

P(X ≥ 4) = 1 - P(X < 4)

= 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

Using the Poisson distribution formula, we can calculate the probabilities of each of these events

P(X = k) = (e^-λ × λ^k) / k!

where λ = 6 and k is the number of typos. Thus,

P(X = 0) = (e^-6 × 6^0) / 0! = e^-6 ≈ 0.0025

P(X = 1) = (e^-6 × 6^1) / 1! = 6e^-6 ≈ 0.015

P(X = 2) = (e^-6 × 6^2) / 2! = 18e^-6 ≈ 0.045

P(X = 3) = (e^-6 × 6^3) / 3! = 36e^-6 ≈ 0.091

Plugging these values into the equation above, we get

P(X ≥ 4) = 1 - (e^-6 + 6e^-6 + 18e^-6 + 36e^-6)

≈ 0.847

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The number of adult tickets purchased that day was 36 if the charge is $3.25 for each adult and $1.50 for each student and 400 people paid $1237

Let the number of adults be x

the number of students be y

Total people = 400

x + y = 400

Total receipts = $1,237

Cost of an adult ticket = $3.25

Cost of a student ticket = $1.50

Cost of x adults tickets = 3.25x

Cost of y student tickets = 1.50y

3.25x + 1.50y = 1237

Multiply the first equation by 1.50

1.50x + 1.50y = 600

Subtract the second and above equation

1.75x = 637

x = 364

364 + y = 400

y = 36

Thus, the number of adult tickets purchased is 36.

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The expression of linear inequalities that represents Tami's earnings is as follows: 6x + 12y ≥ 150.

A linear inequality is a mathematical expression that involves a linear function and a relational operator such as <, >, ≤, ≥, or ≠ and it can be used to compare two expressions or values. It defines a range of values that satisfy inequality.

For the scenario painted above, we see that Tani is meant to earn a minimum of $150 Thus, the greater than or equal to symbol  ≥ should be used for the expression. $6 per hour and $12 per hour are also well represented in the equation.

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Yermin's point is the best approximation of the square root of 0.50.

To know whose point is the best approximation of the square root of 0.50 on a number line. We have the points plotted by Pedro, Lena, Harriet, and Yermin.

Step 1: Calculate the square root of 0.50.
[tex]\sqrt{0.50} = 0.707[/tex]

Step 2: Compare the plotted points to the calculated square root value.
Pedro: Between 0.2 and 0.3
Lena: Between 0.4 and 0.5
Harriet: At 0.5
Yermin: Just to the right of 0.7

Step 3: Determine the closest approximation.
Yermin's point (just to the right of 0.7) is the closest to the calculated value of 0.707.

Your answer: Yermin's point is the best approximation of the square root of 0.50.

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Using the seven square tiles numbered 2, 3, 4, 5, 6, 7, and 8, we can make 1260 different 4-digit numbers.

To create a 4-digit number using these seven square tiles, we have to consider the following:

- The first digit cannot be 2 because then the number would only have three digits.
- We can choose any of the remaining six tiles for the first digit, which means there are 6 choices.
- We can choose any of the seven tiles for the second digit, which means there are 7 choices.
- We can choose any of the remaining six tiles for the third digit, which means there are 6 choices.
- We can choose any of the remaining five tiles for the fourth digit, which means there are 5 choices.

Therefore, the total number of 4-digit numbers we can make is:

6 x 7 x 6 x 5 = 1260

So, using the seven square tiles numbered 2, 3, 4, 5, 6, 7, and 8, we can make 1260 different 4-digit numbers.

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The average time patients waited in 2014 was approximately 76 minutes, calculated by dividing the 2015 waiting time by 0.89 as patients waited 11% less time in 2015.

Let's call the average time patients waited in 2014 as per Wenford Hospital records "x" (in minutes). According to the problem statement, patients waited 11% less time in 2015 compared to 2014, so,

0.89x = 68

Solving for x,

x = 68 / 0.89

x ≈ 76.4

Therefore, the average time patients waited in 2014 was approximately 76 minutes (rounded to the nearest minute).

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The sum of terms 8 to 30 in the arithmetic series is -826.

In an arithmetic series, the nth term is given by the formula an = a1 + (n-1)d, where a1 is the first term and d is the common difference between terms.

We are given that a8 = 1 and a30 = -43. Using the formula above, we can write:

a8 = a1 + 7d = 1 (1)

a30 = a1 + 29d = -43 (2)

Subtracting equation (1) from equation (2), we get:

22d = -44

d = -2

Substituting d = -2 into equation (1) and solving for a1, we get:

a1 = 15

Now we can use the formula for the sum of an arithmetic series to find the sum of terms 8 to 30:

S = (n/2)(a1 + an)

S = (23/2)(15 + (-43))

S = -826

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