use the confidence level and sample data to find a confidence interval for estimating the population μ. round your answer to one decimal place.

a group of 64 randomly selected students have a mean score of 38.6 with a standard deviation of 4.9 on a placement test. what is the 90% confidence interval for the mean score, μ, of all students taking the test?

Answer:

True, because if price changes, dod oes average of money. So if you get 5 bucks per 10 scarfs, you'll get 5.50$ for 11 scarfs

Step-by-step explanation:

Lets just say she gets half of what the actual cost is. In total.

Answer:

This is true

Step-by-step explanation:

Answer:

less than

Step-by-step explanation:

Answer:

\(-\frac{\pi }{10}\) radians

Step-by-step explanation:

Answer:  91 is 70% of 130

Step-by-step explanation:

:)

Answer:

130

Step-by-step explanation:

70% of 130 = 91

Answer:

\(2\)

Step-by-step explanation:

Order of Operations: BPEMDAS

Step 1: Write expression

\(\frac{12/2(3)}{3(3)}\)

Step 2: Evaluate

Answer:

X = 2

Step-by-step explanation:

The name of every month ends in the letter y is the given statement. February is a counterexample to this statement. This is because February does not end with the letter 'y'. So the right  option is (c) February.

What is a counterexample?

In mathematics, a counterexample is an example that opposes or disproves a statement, proposition, or theorem. It is a scenario, an instance, or an example that goes against the given statement.

Therefore, a counterexample demonstrates that the given statement is false or invalid.In this case, the statement is: "The name of every month ends in the letter y." We have to find which of the months listed does not end in "y."February is the only month in the options listed that does not end in the letter "y."

Thus, it is a counterexample to the given statement. Therefore, the correct option is C, February.

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The correct answer is the number of bacteria increases by a factor of 5 each day. I've tooken a quiz on it

The correct answer A that the number of bacteria increases by a factor of 5 each day.

A linear equation is an equation that has the variable of the highest power of 1. The standard form of a linear equation is of the form Ax + B = 0.

It is said that a scientist began a study with a sample of 1,500 bacteria.

He noticed that the number of bacteria in the sample after t days can be modeled by the equation P = 1,500 x 5t.

Here, the equation is given P = 1,500 x 5t.

So, 5t terms represent the number of bacteria increases by 5 bacteria each day.

The correct answer is the number of bacteria increases by a factor of 5 each day.

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Answer:

In order to use the Pythogrean theorem you must first use the points to form a triangle then count the side length in order to get your numbers. It should be 7^2+2^2=c^2 and your answer should be 7.28. This person subtracted the x coordinates from the x coordinate and the y from the y to get their numbers.

Step-by-step explanation:

Answer:

x^2 + 6x + 8

Step-by-step explanation:

area of blue = x^2

area of pink= 2x

area of green=4x

area of brown=8

Adding all the area.

we get : x^2 + 6x +8

Answer:

1. 3) measure of angle 3 and measure of angle 5 are supplementary

2. x=16

The real zeros of P(x) are 1, −1, and −1.

We have,

To factor the given polynomial P(x) = x³ + x² − x − 1, we can use the Rational Root Theorem and synthetic division to find one of its roots, and then factor out the corresponding linear factor.

The constant term of P(x) is −1, and the leading coefficient is 1, so any rational root of P(x) must have a numerator that is a factor of −1 and a denominator that is a factor of 1.

Thus, the possible rational roots of P(x) are ±1 and ±1/1 = ±1.

To check which of these possible roots is actually a root of P(x), we can use synthetic division:

1 | 1   1   -1   -1

 |    1    2    1

 |-------------

 | 1   2    1    0

Since the remainder is 0, we have found that 1 is a root of P(x), and we can write:

P(x) = (x − 1)(x² + 2x + 1)

The quadratic factor can be factored further as (x + 1)².

So,

P(x) = (x − 1)(x + 1)²

The real zeros of P(x) are the values of x that make P(x) equal to 0.

These occur when x = 1, and x = −1 (twice).

Therefore,

The real zeros of P(x) are 1, −1, and −1.

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a) To find the solutions to the equation cos(2x) = 0 on the interval 0 <= x < 2π, we can use the property double-angle identity of cosine that states when the cosine of an angle is zero, the angle must be an odd multiple of π/2.

Therefore, we have two cases to consider: 2x = (2n + 1)π/2, where n is an integer. For the first case, solving 2x = (2n + 1)π/2 for x, we have x = (2n + 1)π/4, where n is an integer. Since we want the solutions within the interval 0 <= x < 2π, we can substitute n = 0, 1, 2, and 3 to find the solutions in that range: x = π/4, 3π/4, 5π/4, and 7π/4. The solutions to cos(2x) = 0 on the interval 0 <= x < 2π are x = π/4, 3π/4, 5π/4, and 7π/4.

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Answer:

y = -26

Step-by-step explanation:

[y = 2 - 2x] + [y = -16 + 2(x - 3)^2​]

y = -16 + 2x - 6^2

y = -16 + 2x - 6^2 + 2 - 2x

y = (+2x - 2x) - 16 - 6^2 + 2

y = -16 - 6^2 + 2

y = -16 - 12 + 2

y = - 28 + 2

y = -26

For the standard normal probability distribution, the area under the probability density function to the left of the mean is 0.5.

This means that half of the distribution lies to the left of the mean, and the other half lies to the right. Since the distribution is symmetric, the area under the curve to the left of the mean is equal to the area under the curve to the right of the mean. Therefore, both areas are 0.5 or 50%.

In a standard normal distribution, the probability density function (PDF) is symmetric about the mean. This means that the area under the curve to the left of the mean is equal to the area under the curve to the right of the mean.

Since the total area under the curve represents the probability of the entire distribution, it sums up to 1. Therefore, when we divide the distribution into two equal parts, each part represents 0.5 or 50% of the total area.

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Answer:

10 cm

Step-by-step explanation:

V =1/3 x π x r² x h

94.25 =1/3 x π x 3² x h

94.25 ÷ 3π =h

so, h= 10 cm

the probability that an admitted applicant is a statistics graduate is approximately 0.818 (or 81.8%) when rounded to one decimal place.

We can use Bayes' theorem to calculate the probability that an admitted applicant is a statistics graduate. Let A represent the event that an applicant is a statistics graduate, and B represent the event that an applicant is admitted. Then, we want to find P(A|B), the probability that an admitted applicant is a statistics graduate.

Using the given information, we can determine the following probabilities:

P(A) = 600/1000 = 0.6

P(B|A) = 0.3

P(B|A') = 0.05

where P(A') represents the complement of event A (i.e., an applicant is not a statistics graduate).

We can now apply Bayes' theorem:

P(A|B) = P(B|A)P(A) / [P(B|A)P(A) + P(B|A')P(A')]

Substituting the given probabilities, we get:

P(A|B) = (0.3)(0.6) / [(0.3)(0.6) + (0.05)(0.4)]

P(A|B) = 0.818

Therefore, the probability that an admitted applicant is a statistics graduate is approximately 0.818 (or 81.8%) when rounded to one decimal place.

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Answer:

She drove 12 hours

5+4+3=12

Step-by-step explanation:

Therefore, the directional derivative of f at P=(1,-1,-2) in the direction of A=3i+2j-6k is -2.

To find the directional derivative of f(x,y,z) at P=(1,-1,-2) in the direction of A=3i+2j-6k, we first need to find the gradient of f at P, which is given by:

grad(f) = ∇f = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k

Here, f(x,y,z) = xy + yz + zx, so we have:

∂f/∂x = y + z

∂f/∂y = x + z

∂f/∂z = x + y

Thus, at P=(1,-1,-2), we have:

∇f(P) = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k

= (y+z)i + (x+z)j + (x+y)k

= (0+(-2))i + (1+(-2))j + (1+0)k

= -2i - 1j + 1k

Next, we need to find the unit vector in the direction of A:

|A| = sqrt(3^2 + 2^2 + (-6)^2) = 7

u = A/|A| = (3/7)i + (2/7)j - (6/7)k

Finally, we can compute the directional derivative of f at P in the direction of A as:

(DAf) | 1(1,-1,-2) = ∇f(P) · u

= (-2i - 1j + 1k) · (3/7)i + (2/7)j - (6/7)k

= -6/7 - 2/7 - 6/7

= -2

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The resultant triangle is shown below. The resulting triangle PQR is the image of the original triangle ABC under the given dilation with center P and scale factor 1/2.

To draw the image of △ ABC triangle under a dilation with center P and scale factor 1/2, follow these steps:

Locate point P: Identify point P, the center of dilation, on the coordinate plane.

Plot the original triangle ABC: Plot the three given points A(0,6), B(-6,0), and C(6,0) to form the original triangle ABC.

Calculate the new coordinates: To find the new coordinates A', B', and C', multiply the x and y coordinates of each point by the scale factor 1/2. For instance, the new coordinates of point A' would be

\((0 \times 1/2, 6 \times 1/2) = (0, 3).\)

Draw the new triangle PQR: Connect the new points A', B', and C' to form the image triangle PQR.

Therefore the resulting triangle PQR is the image of the original triangle ABC under the given dilation with center P and scale factor 1/2. The new triangle will be smaller than the original, with sides reduced by a factor of 1/2.

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Answer:

558 weeks

Step-by-step explanation:

let x = number of weeks

this equation can be derived from the question

2 + 1x = 560

collect like terms

x = 560 - 2

x = 558

Answer:

FG≅LJ

GH≅JK

HF≅KL

Step-by-step explanation:

The required differential equation is x'' + 0.67x' + 2.33x = 0.

The standard differential equation, which we use to represent a damped spring-mass system is written as:

m(d²x/dt²) + c(dx/dt) + kx = 0,

where m is the mass, c is the damping coefficient, k is the spring constant, x is the d²splacement, and the equation has been derived with respect to time t.

In the question, the

mass m = 3 kg,

spring constant k = 7 N/m,

damping constant c = 2N-s/m.

Differentiating x with time t, we take (d²x/dt²) = x'', and (dx/dt) = x'.

Substituting all the values in the standard differential equation, we get:

3x'' + 2x' + 7x = 0,

or, x'' + (2/3)x' + (7/3)x = 0 {Dividing by 3},

or, x'' + 0.67x' + 2.33x = 0, which is the required differential equation.

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The provided question is incorrect. The correct question is:

"Suppose a spring with a spring constant of 7 N/m is horizontal and has one end attached to a wall and the other end attached to a 3 kg mass. Suppose that the friction of the mass with the floor (i.e., the damping constant) is 2 N⋅s/m. Set up a differential equation that describes this system. Let x denote the displacement, in meters, of the mass from its equilibrium position, and give your answer in terms of x, x′, x′′. Assume that positive displacement means the mass is farther from the wall than when the system is at equilibrium."

Answer:0.4948

Step-by-step explanation:

(1+2.46^2) ÷ 2+3.5^2

(1+6.052) ÷ (2+12.25)

7.052/14.25

0.4948

Answer: 15.77

Step-by-step explanation:

To solve for x at an angle, you must first understand what type of angle you are working with. If the angle is a right angle, you can use the Pythagorean theorem to solve for x. If the angle is an obtuse angle, you can use the law of cosines to solve for x. If the angle is an acute angle, you can use the law of sines to solve for x.

To solve for x at a right angle using the Pythagorean theorem, you must have the lengths of the two neighboring sides of the triangle. The sum of the squares of the two sides is equal to the square of the hypotenuse, according to the Pythagorean theorem. By rearranging the components and solving for x, this equation may be used to find x.

To use the law of cosines to solve for x at an obtuse angle, you must know the lengths of all three sides of the triangle. According to the law of cosines, the square of one side's length equals the sum of the squares of the other 2 sides minus twice the product of the other 2 sides multiplied by the cosine of the obtuse angle. You can use this equation to solve for x by rearranging the terms and solving for x.

To apply the law of sines to solve for x in an acute angle, you should first know the lengths of the triangle's two sides as well as the acute angle's measurement. According to the law of sines, the ratio of one side's length to the sine of the opposite angle equals the ratio of the other side's length to the sine of the acute angle. By rearranging the terms and solving for x, you may use this equation to solve for x.

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