A researcher wants to test the hypothesis that on average girls perform better on standardized tests than boys. Since it is true that some boys do better than girls and some girls do better than boys, the researcher must use a two tailed test.
True OR False

Answer:

The answer might be C.(x=0)

Step-by-step explanation:

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Answer:

seven and seven hundred nintey three thousandths

Step-by-step explanation:

Answer: The second one (B)

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Step-by-step explanation:

The times when Hayden made the bracelets in this situation don't matter. All that matters in this problem is how many he made. Since it tells you that he made them for 5 days, you would multiply the number of bracelets by the days, to get the total. (Or write the equation like so: 5*(2+4))

And of course, since they are divided into groups as the last step, we have division.

5*(2+4) /6

The total number of hours spend on the communicating is 11.2 hours each day..

As the stated question-

Average number of waking hour per day = 16.

Communicating hours = 70% of total hours.

Thus,

Communicating hours =  70 x 16 / 100

Communicating hours = 11.2 hours

Thus, the total number of hours spend on the communicating is 11.2 hours.

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Wyatt can husk anywhere from 12 to  18 dozen ears of corn per hour.

To find the possible amount we need to calculate the maximum and minimum time period in hours

Thus, if we need to find the time taken to husk 72 dozen ears of corn, the maximum time would be  ,

=72 / 12 hours

​ =6 hours

And the minimum time would be  used for the does wyatt takes to husk the 72 dozen of corn

=72/18 hours

​=4 hours

Thus, depending on his speed, Wyatt takes anywhere from  4  to 6 hours to husk  72 dozen ears of corn.

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The coordinates of the midpoint are (-1, 3.5) and the distance of AC is 26.25 units.

We know that for a segment whose endpoints are (x₁, y₁) and (x₂, y₂), the coordinates of the midpoint are:

( [x₁ + x₂]/2, [y₁ + y₂]/2)

In this case, the endpoints are:  (3, -9) and (-5, 16), then the midpoint is:

M = ( [3 - 5]/2, [-9 + 16]/2)

M = (-1, 3.5)

The distance between two points (x₁, y₁) and (x₂, y₂) is:

d = √( (x₁ - x₂)^2 + (y₁ - y₂)^2)

So here the distance is:

d = √( (3 + 5)^2 + (-9 - 16)^2) = 26.25 units.

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Answer:

There are 2 stars

Step-by-step explanation:

There are 29 items in total.

As we are given the number of two of the items(nutcrackers and snowmen)we can easily find the number of the last item(stars) by subtracting the number of nutcrackers and snowman by the number of total items.(29)

So the solution would be :

29 - (13 + 14) = number of stars

==> add 13 and 14

29 - 27 = number of stars

==> simplify

2 = number of stars

The number of fish in year 6 will be;

⇒ y = 438.60

What is an expression?

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

The exponential regression equation is,

⇒ y = 14.08 × 2.08ˣ

Where, x is represent the year.

Now,

Since, The exponential regression equation is,

⇒ y = 14.08 × 2.08ˣ

Where, x is represent the year.

Hence, The number of fish in year 6 will be;

⇒ y = 14.08 × 2.08ˣ

⇒ y = 14.08 × 2.08⁶

⇒ y = 14.08 × 31.15

⇒ y = 438.60

Thus, The number of fish in year 6 will be;

⇒ y = 438.60

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The value of the definite integral is 10 after breaking the limits of the integral.

It is defined as the mathematical approach to calculating the smaller parts or components.

It is given that:

The piecewise function is given:

[tex]\rm \int\limits^4_0 {x} \, dx = \int\limits^2_0 {2} \, dx +\int\limits^4_2 {x} \, dx[/tex]

[tex]\rm \int\limits^4_0 {x} \, dx = x^2|^2_0+(1/2)x^2|^4_2[/tex]

= 4 + (1/2)[16 - 4)

= 4 + 6

= 10

Thus, the value of the definite integral is 10 after breaking the limits of the integral.

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Answer:

slope = 1/8

equation: y = 1/8 x

Step-by-step explanation:

Let distance = y.

Let time = x.

y = mx + b

Look at point (1, 8).

Also, the line passes through (0, 0).

slope = m = 1/8

Since it is a direct proportion, b = 0.

Equation: y = 1/8 x

Mathematically, three dwarf hamsters can live in the cage Harriet purchased.

The number of dwarf hamsters that can live in Harriet's cage is determined using mathematical operations, including multiplication and division.

Mathematical operations involve the manipulation of numbers, variables, and mathematical operands to determine mathematical results.

The result of an addition is known as the sum while for a subtraction operation is the difference.

The result of a multiplication operation is known as the product like a dvision operation (also called the quotient).

The height of the cage = 12 inches or 1 foot

The length of the cage = 2¹/₃  feet

The area of the cage = 2¹/₃ ft² (1 x 2¹/₃)

The living space required by each dwarf hamster = ²/₃ ft²

The number of dwarf hamsters that can live in the cage = 3.499 (2¹/₃ ÷ ²/₃).

= 3

Thus, using mathematical operations, we can determine that the highest number of dwarf hamsters that can live in Harriet's cage is three.

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the ratio of blue marvel is 4/3

the ratio of Al marbles to blue is 7/3

The probability that you have seen both the red and green marbles at least once is 0.3104

Given,

1/5 chance of winning a marble of any color in a drawing.

Every draw is independent and has the same probability because the marbles are replaced.

Calculating = 1- the likelihood of not receiving both red and green marbles in any of the four draws yields the same result as P, the probability of receiving both red and green marbles at least once (Q).

The union of the following three examples represents ways to not receive both red and green in all four draws:

(1)

Probability of no red and no green in a draw = 3/5.

Probability of no red and no green in 4 draws,

Q1 = 3/5 * 3/5 * 3/5* 3/5 = 81/54

The probabilities are multiplied as each draw is independent and has the same probabilities.

(2)

The probability of no red and 1 green in 4 draws is given by binomial probability.

= (4 3) (1/5)(3/5)³ = (4 * 3³)/5¹ = 108/5⁴

The probability of no red and 2 green in 4 draws is given by binomial probability.

= (4 2) (1/5)²(3/5)² == (6 * 3²)/54 = 54/5⁴

The probability of no red and 3 green in 4 draws is given by binomial probability.

= (4 3) (1/5)³(3/5) == (4 * 3)/5¹ = 12/5⁴

The probability of no red and 4 green in 4 draws is given by binomial probability.

= (4 4) = 1 * 1/5¹ = 1/5⁴⁴

Q2= sum of all 4 above probabilities.

Q2 (108 +54 + 12+1)/5 = 175/5⁴

(3)

This probability is the same as in Q2. Q3 = 175/5⁴

Therefore the probability of not seeing both red and green at all,

Q = Q1 + Q2 + Q3.

   = 81 +175 + 175/5⁴ = 413/5⁴

Finally, the probability of getting both red and green at least once,

P = 1 - Q

  = 1-413/5⁴

  = (625 431)/625

  = 194/625

  = 0.3104.

The answer is 0.3104.

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Consider a sequence of random variables X1, X2, X3, ⋯ that is defined on an underlying sample space S. For simplicity, let us assume that S is a finite set, so we can write

                                    S = {s₁, s₂,⋯,sₙ }.

Remember that each Xₙ is a function from S to the set of real numbers. Thus, we may write

                       Xₙ (sₐ) = xₙₐ,           for a=1,2,⋯,k.

After this random experiment is performed, one of the sₐ' s will be the outcome of the experiment, and the values of the Xₙ's are known. If sₐ is the outcome of the experiment, we observe the following sequence:

x₁ₐ ,x₂ₐ,x₃ₐ,⋯.

Let [Xₙ] be a sequence of square integral random variables defined on a sample space Omega. We say that [Xₙ] is mean-square convergent (or convergent in mean-square) if and only if there exists a square integral random variable X such that

                                   [tex]\lim_{n \to \infty} E[ X_{n} - X^{2} ][/tex]

Convergence in distribution is in some sense the weakest type of convergence. All it says is that the CDF of Xₙ's converges to the CDF of X as n goes to infinity. It does not require any dependence between the Xₙ's and X. We saw this type of convergence before when we discussed the central limit theorem. To say that Xₙ converges in distribution to X, we write:

                                       Xₙ →d X.

Convergence in probability is stronger than convergence in distribution. In particular, for a sequence X₁, X₂, X₃, ⋯ to converge to a random variable X, we must have that P(|Xₙ−X|≥ϵ) goes to 0 as n→∞, for any ϵ>0. To say that Xₙ converges in probability to X, we write:

                                                Xₙ →p X.

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Answer:

Step-by-step explanation:

Answer:

(10) x=1, y=-3

(12) x=3, y=22

(13) x= 6, y= 7

Step-by-step explanation:

[10] 6x -9 =-3x

6x+3x=9

9x=9

x=9,

The values that represent the solution to [tex]cos (\frac{\pi }{4} - x) = \frac{\sqrt{2} }{2} sin x[/tex] ,

x ∈ [0, 2·π) are:

              [tex]x = (\frac{\pi }{2} , \frac{3\pi }{2} )[/tex]

Sine and Cosine are trigonometric functions of an angle. The sine and cosine of an acute angle are defined in the context of a right triangle: for the specified angle, its sine is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle (the hypotenuse), and the cosine is the ratio of the length of the adjacent leg to that of the hypotenuse. For an angle θ the sine and cosine functions are denoted simply as sinθ and cos θ.

Once such a triangle is chosen, the sine of the angle is equal to the length of the opposite side, divided by the length of the hypotenuse:

Sin (α) = [tex]\frac{opposite}{Hypotenuse}[/tex]  and  Cos (α) = [tex]\frac{Adjacent}{Hypotenues}[/tex]

Given in the question:

The given function is presented as follows;

[tex]cos (\frac{\pi }{4} - x) = \frac{\sqrt{2} }{2} sin x[/tex]

Where;

x ∈ [0, 2·π)

We have;

cos(A - B) = cos(A)·cos(B) - sin(A)·sin(B)

[tex]cos (\frac{\pi }{4} ) = \frac{\sqrt{2} }{2} , sin( \frac{\pi }{4)} =\frac{\sqrt{2} }{2}[/tex]

Which gives

[tex]cos(\frac{\pi }{4} - x ) = \frac{\sqrt{2} }{2} cos x +\frac{\sqrt{2} }{2} sinx[/tex]

Therefore, we get;

[tex]\frac{\sqrt{2} }{2} cos x = \frac{\sqrt{2} }{2} sinx - \frac{\sqrt{2} }{2} = 0[/tex]

cos x = 0

[tex]cos (\frac{\pi }{2} ) =0[/tex]

Therefore;

The possible solutions (x-values) to cos x = 0 are;

                             [tex]0\leq \frac{n.\pi }{2} \leq 2\pi[/tex]

Where;

n = 1, 3, 5, ...

x ∈ [0, 2·π)

We get;   [tex]x = \frac{\pi }{2} or \frac{3\pi }{2}[/tex]

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The area of the piece of land is 849 ft².

The area of the piece of land

=22 x 30 x [tex]\frac{1}{2}[/tex] x 2 + (30-12) x 21 x [tex]\frac{1}{2}[/tex]

=660 + 18 × 21×[tex]\frac{1}{2}[/tex]

=660 + 189

=849 ft²

The area is the amount that expresses the volume of a place on the plane or on a curved surface. The location of an aircraft region or aircraft place refers back to the area of a shape or planar lamina, at the same time as floor region refers to the location of an open floor or the boundary of a three-dimensional object. the area may be understood as the amount of fabric with a given thickness that could be essential to style a model of the shape, or the amount of paint vital to cowl the surface with a single coat. it's miles the 2-dimensional analog of the length of a curve (a one-dimensional concept) or the extent of a strong (a three-dimensional concept).

The place of a shape can be measured by way of evaluating the form to squares of a hard and fast size. inside the global gadget of devices (SI), the usual unit of the vicinity is the rectangular meter (written as m²), that's the location of a rectangle whose sides are one meter long. A shape with a place of three rectangular meters would have the equal region as 3 such squares. In mathematics, the unit square is described to have placed one, and the location of another shape or floor is a dimensionless real variety.

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Using the probability formula, the number of diamonds needed to make the game fair is determined by option (B) 6.

Assume that there are a finite number of elementary events that are all equally likely to occur in the sample space of the experiment under consideration.

So let's say we want to determine the likelihood of an occurrence E.

So, the value of diamonds has to be:

The expected number of points that one can lose should equal the expected number of points that one can acquire for the game to be fair.

So, in order for the game to be fair, we need:

Expected points earned by selecting a diamond minus anticipated points lost by selecting a diamond.

If we choose a diamond, let's say we obtain P points.

Let X represent the random variable that records the player's score for each individual draw.

13/52 = 1/4 is the probability of selecting a diamond from a deck of 52 cards.

This is due to the fact that there are only 13 ways to choose a diamond out of the total 52 ways to choose a card.

So, the likelihood of receiving P points is 1/4, or 0.25.

The odds of not getting a diamond equal 1 minus the odds of getting a diamond equal 1 minus 0.25, which equals 0.75.

The points we receive when we don't acquire a diamond are -2. (negative since we actually lose it, so we can take it as a negative gain).

E(X) = expectation of X = 0 is therefore required (since the negative points of loosing would equate to positive point of winning, thus, making the expected overall score 0, thus, keeping it a fair game.)

We now have:

When choosing a non-diamond card, P(X = -2) equals 0.75; when choosing a diamond card, P(X = p) equals 0.25.

There are only possible values for X.

The result is:

E(X) = -2*p(X=-2)+p*p(X=p)

E(X) = -2*0.75+p*0.25

0=-1.5+0.25p

p = 1.5/0.25 = 6

Therefore, using the probability formula, the number of diamonds needed to make the game fair is determined by option (B) 6.

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Complete question:
You pick a card at random from an ordinary deck of 52 cards. If the card is a diamond, you get some points; if not, you lose 2 points. What value for the diamonds would make the game fair?

a) 12

b) 6

c) 24

d) 3

Answer:

B-6

Step-by-step explanation:

Round to the closest whole percent, 23% of the participants in the event are under the age of 20.

For the stated question-

Total number of people = 30 + 50 + 40 + 10

Total number of people = 130

Number of participants are under the age of 20 = 30.

Let 'x' be the % of the participants are under the age of 20.

Thus,

x% of 130 = 30

x = 30 x 100 / 130

x = 23.07

x = 23%

Thus, round to the closest whole percent, 23% of the participants in the event are under age of 20.

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Answer:

a = 40

Step-by-step explanation:

Answer: 3 boxes

Step-by-step explanation:

6 x 5 = 30

36(closest multiple of 12 to 30) divided by 12 = 3

Answer:

2.5 boxes

Step-by-step explanation:

So, if each friend wants 5 cookies then you would multiply 6*5 which =30. Then you would divide 30 by 12 boxes which = 2.5. So you need 2.5 boxes or 3 if they need to be whole.

Answer:

Step-by-step explanation:

You need to set the problem up like this:

         back in for all the 'x's

        should be 20

(767.47, 777.13) is the confidence interval in the given situation.

A confidence interval is a range of estimates for an unknown parameter in frequentist statistics.

The 95% confidence level is the most popular, but other levels, like 90% or 99%, are occasionally used when computing confidence intervals.

[tex]\begin{aligned}&\mathrm{CI}=(772.3-4.83,772.3+4.83) \\&\mathrm{CI}=(767.47,777.13)\end{aligned}[/tex]

Therefore, (767.47, 777.13) is the confidence interval in the given situation.

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Given that a population proportion represents the proportion of successes within a population and that there must be two outcomes (categories), success and failure, a confidence interval about a population proportion only makes sense for qualitative variables.

A confidence interval (CI) is a range of estimates for an unknown parameter in frequentist statistics.

The 95% confidence level is the most popular, however other levels, such 90% or 99%, are occasionally used when computing confidence intervals.

The percentage of related CIs over the long run that include the parameter's actual value is represented by the confidence level. For instance, 95% of all intervals calculated at the 95% confidence level should include the parameter's actual value.

The sample size, sample variability, and confidence level are all variables that have an impact on the CI's width

Assuming all other factors remained constant, a larger sample would result in a narrower confidence range. Additionally, more sample variability results in a wider range of.

Hence,  a population proportion represents the proportion of successes within a population and that there must be two outcomes (categories), success and failure, a confidence interval about a population proportion only makes sense for qualitative variables.

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