A genetic experiment involving peas yielded one sample offspring consisting of 412 green peas and 125 yellow peas. Use a 0.05 significance level to test the most under the same circumstances, 20% of offspring peas will be yellow. Identify the hypothesis, alternative hypothesis et statistics, P-value, conclusion about the null hypothesis and final conclusion that adds the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
What are the null and alternative hypotheses?
A H_o: p≠0.26 H_1:p=0.26
B H_o: p≠0.26 H_1:p>0.26
C H_o: p=0.20 H_1:p<0.26
D H_o: p=0.26 H_1:p>0.26
E H_o: p=0.26 H_1:p≠0.26
F H_o: p≠0.20 H_1:p<0.26
What is the test statistic?
z = _______
(Round local places as needed)
What is the P-value?
P-value = _________

Answer:

3328 cubic centimeters

Step-by-step explanation:

volume of pyramid equation:

V=(lwh)/3

V = (12·26·32) / 3

V =3328

Answer:

The answer is

[tex]9984 cm {}^{3} [/tex]

Step-by-step explanation:

The way i solved this was by using the formula to volume. I also am doing this but for me it is a bit easier. The simple formula is Width x Length x Height. Since i already have the numbers, it is easier to plug in the numbers

Answer:

G

Step-by-step explanation:

out of a total of 280 spinners as the overall.

3/40 were defective

280 * 3/40 = 21

Answer:

G

Step-by-step explanation:

For every 40 spinners 3 are defective

Divide amount made by 40 for numbers of groups of 40

280 ÷ 40 = 7 , then

7 × 3 = 21 ← likely defective spinners → G

The critical points function relative extrema and saddle points.

(a) f(x, y) = x - x2 - y2 =f(x, y) = 0: x - x² - y²= 0

(b) f(x, y) = (x + 1)(y – 2)=: x + 1 = 0 and y - 2 = 0.

(c) f(x, y) = sin(xy)= cos(xy),

(d) f(x, y) = xy(x - 1)= (0, 0) and (1, y)

(a) For the function f(x, y) = x - x² - y²

To find the critical points, to find where the gradient is zero or undefined. The gradient of f(x, y) is given by (∂f/∂x, ∂f/∂y):

∂f/∂x = 1 - 2x

∂f/∂y = -2y

Setting both partial derivatives to zero,

1 - 2x = 0 -> x = 1/2

-2y = 0 -> y = 0

The only critical point is (1/2, 0).

To determine the nature of the critical point, examine the second-order partial derivatives:

∂²f/∂x² = -2

∂²f/∂y² = -2

∂²f/∂x∂y = 0

The determinant of the Hessian matrix is Δ = ∂²f/∂x² * ∂²f/∂y² - (∂²f/∂x∂y)² = (-2)(-2) - (0)² = 4.

Since Δ > 0 and ∂²f/∂x² = -2 < 0, the critical point (1/2, 0) is a local maximum.

To sketch the level sets,  set f(x, y) to different constant values and plot the corresponding curves. For example:

f(x, y) = -1: x - x² - y² = -1

This equation represents a circle with radius 1 centered at (1/2, 0).

f(x, y) = 0: x - x² - y² = 0

This equation represents a parabolic shape that opens downward.

(b) For the function f(x, y) = (x + 1)(y - 2):

To find the critical points, we set both partial derivatives to zero:

∂f/∂x = y - 2 = 0 -> y = 2

∂f/∂y = x + 1 = 0 -> x = -1

The only critical point is (-1, 2).

To determine the nature of the critical point, we can examine the second-order partial derivatives:

∂²f/∂x² = 0

∂²f/∂y² = 0

∂²f/∂x∂y = 1

Since the second-order partial derivatives are all zero, we cannot determine the nature of the critical point based on them. We need further analysis.

To sketch the level sets,  set f(x, y) to different constant values and plot the corresponding curves. For example:

f(x, y) = 0: (x + 1)(y - 2) = 0

This equation represents two lines: x + 1 = 0 and y - 2 = 0.

(c) For the function f(x, y) = sin(xy):

To find the critical points, both partial derivatives to zero:

∂f/∂x = ycos(xy) = 0 -> y = 0 or cos(xy) = 0

∂f/∂y = xcos(xy) = 0 -> x = 0 or cos(xy) = 0

From y = 0 or x = 0, the critical points (0, 0).

When cos(xy) = 0,  xy = (2n + 1)π/2 for n being an integer. In this case,  infinitely many critical points.

To determine the nature of the critical points, we can examine the second-order partial derivatives:

∂²f/∂x² = -y²sin(xy)

∂²f/∂y² = -x²sin(xy)

∂²f/∂x∂y = (1 - xy)cos(xy)

Since the second-order partial derivatives involve the trigonometric functions sin(xy) and cos(xy), it is challenging to determine the nature of the critical points without further analysis.

To sketch the level set f(x, y) to different constant values and plot the corresponding curves.

(d) For the function f(x, y) = xy(x - 1):

To find the critical points, both partial derivatives to zero:

∂f/∂x = y(x - 1) + xy = 0 -> y(x - 1 + x) = 0 -> y(2x - 1) = 0

∂f/∂y = x(x - 1) = 0

From y(2x - 1) = 0,  y = 0 or 2x - 1 = 0. This gives us the critical points (0, 0) and (1/2, y) for any y.

From x(x - 1) = 0,  x = 0 or x = 1. These values correspond to the critical points (0, 0) and (1, y) for any y.

To determine the nature of the critical points,  examine the second-order partial derivatives:

∂²f/∂x² = 2y

∂²f/∂y² = 0

∂²f/∂x∂y = 2x - 1

For the critical point (0, 0), the second-order partial derivatives are ∂²f/∂x² = 0, ∂²f/∂y² = 0, and ∂²f/∂x∂y = -1. Based on the second partial derivative test, this critical point is a saddle point.

For the critical points (1, y) and (1/2, y) where y can be any value, the second-order partial derivatives are ∂²f/∂x² = 2y, ∂²f/∂y² = 0, and ∂²f/∂x∂y = 1. The nature of these critical points depends on the value of y.

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The box and whisker plot of the prices of some DVDs:

Minimum: 9.99

First Quartile: 12.99

Median: 15.99

Third Quartile: 19.99

Maximum: 26.99

The box and whisker plot shows that the median price of a DVD is $15.99. The prices range from $9.99 to $26.99. There are two outliers, one at $9.99 and one at $26.99.

The box and whisker plot can be used to identify the distribution of the data. In this case, the data is slightly skewed to the right, meaning that there are more DVDs priced at the lower end of the range than at the higher end.

The box and whisker plot can also be used to compare different sets of data. For example, we could compare the prices of DVDs from different stores or from different years.

Overall, the box and whisker plot is a useful tool for visualizing and summarizing data.

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Probability refers to the measure of the likelihood that a particular event will occur. It is represented as a value between 0 and 1, where 0 indicates an impossible event and 1 indicates a certain event.

The probability of not rolling a 2 on a number cube with sides numbered 1 through 6 is 5/6.

Here's why: When we roll a number cube with sides numbered 1 through 6, there are six possible outcomes, each with an equal probability of 1/6:1, 2, 3, 4, 5, 6.The probability of rolling a 2 is 1/6, which means there is only one way to roll a 2 out of the six possible outcomes. The probability of not rolling a 2 is the probability of rolling any of the other five possible outcomes. Each of these outcomes has an equal probability of 1/6. Therefore, the probability of not rolling a 2 is:1 - (1/6) = 5/6. Answer: b. 5/6.

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Given that the number cube has sides numbered 1 through 6. The probability of rolling a 2 is 1/6. The probability of not rolling a 2 on a number cube with sides numbered 1 through 6 is 5/6.

The probability of rolling any of the numbers 1, 3, 4, 5, or 6 is also 1/6 each.

The sum of the probabilities of all possible outcomes is 1.

The probability of an event happening is defined as the number of ways the event can occur, divided by the total number of possible outcomes.

The total number of possible outcomes is 6 (the numbers 1 through 6).

Thus, if the probability of rolling a 2 is 1/6, then the probability of not rolling a 2 is 1 - 1/6 = 5/6.

Therefore, the correct option is b. 5/6.

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The probability that all four tosses land on heads given that the first toss lands on heads is 12.5%.

Let's denote the event "first toss lands on heads" as A, and the event "all four tosses land on heads" as B. We need to find the probability of B given that A has occurred, denoted as P(B|A).

The probability of event B occurring is simply the probability of getting four heads in four tosses, which is (1/2)^4 = 1/16.

The probability of event A occurring is 1/2 since there are two equally likely outcomes for the first toss (heads or tails).

To find the conditional probability P(B|A), we use the formula:

P(B|A) = P(A ∩ B) / P(A)

Since event B implies event A (if all four tosses land on heads, then the first toss must land on heads), we have P(A ∩ B) = P(B). Therefore:

P(B|A) = P(B) / P(A) = (1/16) / (1/2) = 1/8.

Converting the probability to a percentage, we have:

P(B|A) = (1/8) × 100 = 12.5%.

Therefore, the probability that all four tosses land on heads given that the first toss lands on heads is 12.5%.

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1.

a. 8 x 7 x 1/2 = 56 x 1/2 = 28

b. 24 x 14 x 1/2 = 336 x 1/2 = 168

c. 5 x 12 x 1/2 = 60 x 1/2 = 30

d. 4 x 4.8 x 1/2 = 19.2 x 1/2 = 9.6

2.

two triangles: 2(4 x 8 x 1/2) = 4 x 8 = 32

rectangle: 15 x 8 = 120

total area: 32 + 120 = 152

Mario has gross biweekly earnings of $784.21. By claiming 1 more withholding allowance, Mario would have $13 more in his take home pay. so the correct answer is b. 4.

By claiming 1 more withholding allowance, Mario would have $13 more in his take-home pay. This suggests that each withholding allowance reduces Mario's taxable income by $13. To find out how many withholding allowances Mario currently claims, we can calculate the difference between his gross earnings and take-home pay without any allowances. If we subtract $13 from the take-home pay ($784.21 - $13 = $771.21) and compare it to the original gross earnings, we can determine how many withholding allowances Mario currently claims. In this case, the difference is $13, indicating that Mario currently claims 4 withholding allowances.

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Answer:

XY would also be 7 centimeters which is answer D.

Step-by-step explanation:

This is a parallelogram, meaning that the adjacent sides are congruent. As well, the triangles making up the figure are congruent, so it makes sense that XY would also equal 7 centimeters.

Answer:

100degrees

Step-by-step explanation:

Find the diagram attached

From the diagram

<PQT =<TRS = 40

Since the triangle STR is isosceles, here,

<TSR =<TRS = 40

Also the sum of angle in a triangle is 189, hence:

Arc SR+<TSR +<TRS = 18₩

ArcSR +40+40=180

ArcSR +80=180

ArcSR = 180-80

ArcSR = 100degrees

Hence the measure of SR is 100degrees

The measure of arc SR is 100°. The correct option is the third option - 100°

From the question, we are to determine the measure of arc SR.

The measure of arc SR = <STR

Now, we will determine the measure of <STR

In the diagram, T is the center of the circle.

∴ TP and TQ are radii.

Then, we can conclude that ΔPQT is an isosceles triangle.

Recall: Base angles of an isosceles triangle are equal.

∴ <PQT = <TPQ = 40°

Now, consider ΔPQT

<QTP + <TPQ + <PQT = 180° (Sum of angles in a triangle)

<QTP + 40° + 40° = 180°

<QTP + 80° = 180°

<QTP = 180° - 80°

<QTP = 100°

Also, in the diagram, <QTP and <STR are vertically opposite angles

NOTE: Vertically opposite angles are equal

That is, <QTP = <STR

∴ <STR = 100°

Hence, the measure of arc SR is 100°. The correct option is the third option 100°

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Here is the complete and correct question:

Line PR and Line QS are diameters of circle T. What is the measure of Arc SR?

50°

80°

100°

120°

Please find the attached image

The exact value of the integral [tex]\int\limits^0_{\pi/6}[/tex] 2/cos²θ dθ is √3/3 using the Fundamental Theorem of Calculus.

To calculate the integral ∫[0 to π/6] 2/cos²θ dθ using the Fundamental Theorem of Calculus, we need to find the antiderivative of the integrand and evaluate it at the upper and lower limits of integration.

The antiderivative of 2/cos²θ is tan(θ), so applying the Fundamental Theorem of Calculus:

[tex]\int\limits^0_{\pi/6}[/tex] 2/cos²θ dθ = [tan(θ)] evaluated from θ = 0 to θ = π/6

Substituting the upper and lower limits of integration:

= tan(π/6) - tan(0)

Since tan(0) = 0, we have:

= tan(π/6)

Using the value of tan(π/6) = √3/3, the exact value of the integral is √3/3.

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The question is -

Use the Fundamental Theorem to calculate the following integral exactly:

[tex]\int\limits^0_{\pi/6}[/tex] 2/cos²∅ d∅ = _______

Quotient: The result of dividing two numbers

Explanation: Just some simple dividing and rounding

Quotient - 102.756098

Rounding - 102.76

Answer: 102.76

The given problem is about hypothesis testing. The sample size is 40, and the proportion of students reporting their studies in the new situation (online) is the same as in the face-to-face context is 46%.

1A. The expected numbers of university students who reported the same learning experience in online and face-to-face contexts are 18.4.

1B. The degrees of freedom associated with this hypothesis test is 39.

1C. The value of the test statistic associated with this hypothesis test is approximately 0.518.

Here, the null hypothesis is H0: p = 0.46 and the alternative hypothesis is Ha: p ≠ 0.46, where p is the proportion of university students reporting the same learning experience in online and face-to-face contexts.

Here, we are interested in testing whether the proportion of students reporting an equal preference for online and face-to-face learning has changed due to the Covid-19 pandemic.

1A. Assuming the hypothesized value holds, the expected numbers of university students who reported the same learning experience in online and face-to-face contexts are 0.46 × 40 = 18.4.

1B. The degrees of freedom associated with this hypothesis test is (n - 1) where n is the sample size.

Here, n = 40.

Hence, the degrees of freedom will be 40 - 1 = 39.

1C. The value of the test statistic associated with this hypothesis test can be calculated as follows:

z = (X - μ) / σ, where X = 20,

μ = np

μ = 18.4, and

σ = √(npq)

σ = √(40 × 0.46 × 0.54)

σ ≈ 3.09.

z = (20 - 18.4) / 3.09

z ≈ 0.518

So, the value of the test statistic associated with this hypothesis test is approximately 0.518.

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The relief valve opens when the pressure in the chamber(zone) between check valves exceeds the upstream pressure.

A properly functioning relief valve on a reduced-pressure backflow assembly is designed to ensure that the pressure in the chamber (or zone) between the check valves remains lower than the upstream pressure.

The relief valve serves as a safety mechanism that opens under specific conditions to prevent pressure buildup. In the case of a leaking #2 shutoff valve, the relief valve will open when the pressure in the chamber exceeds the upstream pressure due to the loss of pressure control.

Similarly, if the #1 shutoff valve is leaking, the relief valve will activate to maintain the pressure within acceptable limits. Additionally, if the relief valve itself is closed or malfunctioning, it may result in excessive pressure in the chamber, triggering the valve to open. Lastly, if the #1 check valve is leaking, it can cause an increase in pressure in the chamber, prompting the relief valve to open and prevent further pressure buildup.

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Step-by-step explanation:

You can apply cosinus theory for finding are

Area=cos40°3.4(ft)*2.7(ft)/2 like thia

Answer:

Should be 9 centimeters.

Step-by-step explanation:

If "f" is function with derivative as f'(x) = √(x³ + 1), then length of graph of y = f(x) from x = 0 to x = 1.5 is (b) 2.497.

To find the length of the graph of y = f(x) from x = 0 to x = 1.5, we use the arc-length formula for a function y = f(x):

Length = ∫ᵇₐ√(1 + [f'(x)]²) dx,

Given the derivative : f'(x) = √(x³ + 1), we substitute it into the arc-length formula:

Length = [tex]\int\limits^{1.5}_{0}[/tex] √(1 + (√(x³ + 1))²) dx,

Simplifying the expression inside the square root:

We get,

Length = [tex]\int\limits^{1.5}_{0}[/tex] √(1 + x³ + 1) dx

= [tex]\int\limits^{1.5}_{0}[/tex]√(x³ + 2) dx

= 2.497.

Therefore, the correct option is (b).

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The given question is incomplete, the complete question is

Let f be a function with derivative given by f'(x) = √(x³ + 1). What is the length of the graph of y = f(x) from x = 0 to x = 1.5?

(a) 4.266

(b) 2.497

(c) 2.278

(d) 1.976

The probability that an individual distance is greater than 211.80 cm is 0.1292.

The problem statement is:

The overhead reach distances of adult females are normally distributed with mean of 202.5 cm and standard deviation of 8.3 cm Find the probability that an individual distance is greater than 211.80 cm.

We need to find the z-score first as follows:$\begin{aligned}z&=\frac{x-\mu}{\sigma} \\z&=\frac{211.80-202.5}{8.3} \\z&=1.122\end{aligned}$

Using the standard normal table,

The standard normal distribution table is a compilation of areas from the standard normal distribution, more commonly known as a bell curve, which provides the area of the region located under the bell curve and to the left of a given z-score to represent probabilities of occurrence in a given population.

The probability is given by:P(Z > 1.122) = 0.1292

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The given information is Population Mean (µ) is 202.5 cm

Standard deviation (σ) is 8.3 cm

The distance we have to find is x = 211.80 cm.

Hence, the probability that an individual distance is greater than 211.80 cm is 0.1314.

The formula used for finding the probability is: Now we need to find z score to use the standard normal distribution tables. The formula for finding z score is:

z = (x - µ) / σ

Substitute the values in the above formula, we get

z = (211.80 - 202.5) / 8.3

z = 1.12

Now use this z value to look up in the standard normal distribution tables to find the probability.

P(z > 1.12) = 1 - P(z < 1.12)

From standard normal distribution tables, the probability of P(z < 1.12) is 0.8686.

Therefore, P(z > 1.12) = 1 - P(z < 1.12)

= 1 - 0.8686

= 0.1314

Hence, the probability that an individual distance is greater than 211.80 cm is 0.1314.

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Answer:

61

Step-by-step explanation:

add the numbers up and divide them by 7

Answer:

Step-by-step explanation:

added up the data = 427

divide by the number of days = 427/7 = 61

The half-life of the radioactive goo is 48 minutes. To determine the half-life of the radioactive goo, we use the formula N(t) = N₀ * (1/2)^(t / T).

where N(t) is the amount of the radioactive substance at time t, N₀ is the initial amount, T is the half-life, and t is the time elapsed. Given N₀ = 120 grams and N(240) = 3.75 grams, we substitute these values into the formula and solve for T. Plugging in the given values, we have: 3.75 = 120 * (1/2)^(240 / T)

To find the half-life T, we need to isolate it on one side of the equation. We can begin by dividing both sides of the equation by 120:

3.75 / 120 = (1/2)^(240 / T)

0.03125 = (1/2)^(240 / T)

Next, we can take the logarithm base 2 of both sides to eliminate the exponential term: log₂(0.03125) = log₂[(1/2)^(240 / T)]

-5 = (240 / T) * log₂(1/2)

Simplifying further, we know that log₂(1/2) is equal to -1: -5 = (240 / T) * (-1)

To solve for T, we can multiply both sides by -T/240: 5T/240 = 1

Multiplying both sides by 240/5, we find: T = 48

Therefore, the half-life of the radioactive goo is 48 minutes.

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Answer:

Step-by-step explanation:

Answer:

44 is the awnser

Step-by-step explanation:

becuase if you were to look at the first persons work it is correct showing him solving the equasion and witch it is not 44

Answer:Answer: (a) = P = 2(x + 2x-5) = 2(3x-5) = 6x - 10.

Step-by-step explanation:

Part a : Polynomial represents the perimeter of the garden is 6W-10

Part b : Polynomial represents the area of the garden is [tex]2w^{2}-5w[/tex]

Part c : Perimeter = 38 feet and area = 88 feet square

A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables

A perimeter is a closed path that encompasses, surrounds

Area is the quantity that expresses the extent of a region on the plane or on a curved surface

Given,

Length is five less than two times the width

Consider

L is the length and W is the width

Then,

L=2W-5

Part a

Perimeter = 2(L+W)

Substitute the value of L

P=2(L+W)

P=2((2W-5)+W)

P=2(2W-5+W)

P=2(3W-5)

Perimeter= 6W-10

Part b

Area of the garden = L×W

Area = (2W-5)W

Area = [tex]2w^{2}-5w[/tex]

Part c

Given width W = 8 feet

Perimeter = 6W-10

P=6×8-10

Perimeter =38 feet

Area = [tex]2w^{2}-5w[/tex]

Area=[tex]2(8^{2})-5(8)[/tex]

Area = 88 feet square

Hence,

Part a : Polynomial represents the perimeter of the garden is 6W-10

Part b : Polynomial represents the area of the garden is [tex]2w^{2}-5w[/tex]

Part c : Perimeter = 38 feet and area = 88 feet square

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Answer: $42,000

Step-by-step explanation:

Simple Interest: I=prt

I= (6,000)(7/4)(4)

I= 42,000

Answer:

mean: 3.3

median: 3

mode: 3

range: 5

Q1 = 2

Q3 = 4

IQR = 2

Step-by-step explanation:

Answer:

3n + 21

Step-by-step explanation:

3(n + 7)

3n + 21

Answer:

Step-by-step explanation:

Focus of parabola [tex](y-2)^2 = 4(x+3)[/tex] is (-2 , 2) .

Correct option is C .

Given, Equation of parabola  [tex](y-2)^2 = 4(x+3)[/tex]

Focus of parabola :

Standard equation of parabola : (y - k)² = 4a(x - h)

Axis of parabola : y = k

Vertex of parabola : (h, k)

Focus of parabola : (h + a, k)

Compare the equation of parabola with standard equation.

(y - k)² = 4a(x - h)

[tex](y-2)^2 = 4(x+3)[/tex]

k = 2

a = 1

h = -3

So focus of parabola:  (h + a, k).

-3 + 1 , 2

Focus of parabola = -2 , 2

Hence the correct option is C .

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