At a cost of $8 per pound, how many ounces of barley can be bought for $3.70?

Answer:

1.29 x [tex]10^{9}[/tex]

Step-by-step explanation:

{(1.8 x 10^-6) x (4.3 x 10^8)}  / [(8 x 10^-5) x (7.5 x 10^-3)]

[(1.8 x 4.3)x(10^-6 x 8)] / [(8 x 7.5) x(10^-5 x 10^ -3)]

(.129 x 10^[2-(-8)]

.129 x 10^10

1.29 x 10^-1 x 10^10

1.29 10^9

The probability that the first student chosen will be a boy and the second will be a girl is 0.28.

Probability is also called the chance because if you flip a coin then the probability of coming head and tail is nothing but chances that either head will appear or not.

Total students = 9

Number of girls = 4

Number of boys = 5

Probability of choosing first boy = Number of boys / total students

⇒ 5/9

Now since 1 boy extract so remaining students = 9 - 1 = 8

Probability of choosing second girl = Number of girls / Total students

⇒ 4//8

Probability of A and B = P(A) × P(B)

Therefore,

The probability that the first student chosen will be a boy and the second will be a girl = (5/9) × (4/8) = 0.28

Hence "The probability that the first student chosen will be a boy and the second will be a girl is 0.28".

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The elements that will be coming in the set D∪E will be {12, 14, 15, 16, 17}.

A set may be defined as a collection of letters or numbers that are written in order to depict a certain value or entity. A set is always represented by a capital letter symbol and is always written in curly brackets { }. Union of two sets may be defined as a new set which has the collection of all the elements of the two individual sets. Union of two sets is represented by the symbol '∪'. Union of two Indvidual sets, set A and set b is written as A∪B.

Now, according to the question set D is = {12, 15, 17} and set E is = {12, 14, 15, 16}.

The union of the two sets contains all the elements of the sets D and E.

Thus, D∪E will be given by {12, 14, 15, 16, 17}.

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Answer: $2.50

Step-by-step explanation: To find the tax rate, you need to divide 1.50 by 30

1.5/30 = .05, which is the percent taxed. To find the tax amount for a $50 item, you just need to multiply $50 by the tax rate, which is .05, to get $2.50

4 x + 8 = negative 24 , 9 x + 18 = negative 24 are  equations have the same value of x .

What in mathematics is a linear equation?

A - 4 x + 8 = negative 24

E - 4 x = negative 32

ARE CORRECT ON EDGE.

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The complete question is-

Which equations have the same value of x as Two-thirds (6 x + 12) = negative 24? Select two options.

4 x + 8 = negative 24

9 x + 18 = negative 24

4 x = negative 16

StartFraction 18 x + 36 over 2 EndFraction = negative 24

4 x = negative 32

Answer: 29/21

Quotient: 1

Reminder: 8

Step-by-step explanation:

Add up the amount of 7th graders playing instruments (not including drums) total and put that number under the amount of 8th graders playing a instrument.

Use the 1st  digit 2 from dividend 29

21 & 29 squared root

Since 2 is less than 21, use the next digit 9 from dividend 29 and add 0 to the quotient. Find the closest multiple of 21 to 29.

1×21=21 is the nearest. Now subtract 21 from 29 to get reminder 8. Add 1 to quotient.

Since 8 is less than 21, stop the division. The reminder is 8. The topmost line 01 is the quotient. Remove all zeros at the start of the quotient to get the actual quotient 1.

Answer:
The cost that Kaleb would spend at McDonalds if he ordered 3 McGriddles would be 11.50.

Step-by-step explanation:

We know this is the answer because, when solving for c(3), we would get c(3)=2.3(3)+2.50=11.50.

Answer:

Minimum = (-3, -2)

Step-by-step explanation:

Standard form of a quadratic function:

[tex]f(x)=ax^2+bx+c[/tex]

If a > 0 the parabola opens upwards and the curve has a minimum point.

If a < 0 the parabola opens downwards and curve has a maximum point.

Given function:

[tex]f(x)=x^2+6x+7[/tex]

As a > 0, the parabola opens upwards and so the curve has a minimum point.

The minimum/maximum point of a quadratic function is its vertex.

Vertex form of a quadratic function:

[tex]f(x)=(x-h)^2+k[/tex]

Where (h, k) is the vertex.

To rewrite the given function in vertex form, complete the square.

Add and subtract the square of half the coefficient of the term in x:

[tex]\implies f(x)=x^2+6x+7 +\left(\dfrac{6}{2}\right)^2-\left(\dfrac{6}{2}\right)^2[/tex]

[tex]\implies f(x)=x^2+6x+7 +9-9[/tex]

[tex]\implies f(x)=x^2+6x+9+(7 -9)[/tex]

[tex]\implies f(x)=x^2+6x+9-2[/tex]

Factor the perfect square trinomial formed by x²+6x+9:

[tex]\implies f(x)=(x+3)^2-2[/tex]

Compare with the vertex form:

Therefore, the vertex is (-3, -2) and so the minimum value of the given function is (-3, -2).

Answer:

23% or 23/100

Step-by-step explanation:

Since there are 23 in 100 is is 23/100 or 23%

Answer:

4.348%

Step-by-step explanation:

blue marbles = 19

yellow marbles = 23

probably of getting yellow = 23/100

= 4.34782608696

then you round then number to three decimal places and there you go

Answer:

(-6, 7)

Step-by-step explanation:

90 degrees counterclockwise / 270 degrees clockwise:

(x,y) -> (-y,x)
180 degrees counterclockwise / 180 degrees clockwise:

(x,y) -> (-x,-y)

270 degrees counterclockwise / 90 degrees clockwise:

(x,y) -> (y,-x)

Answer:

b=–2

Step-by-step explanation:

we've got:

(3+bx)⁵===> b⁵x⁵+15b⁴x⁴+90b³x³+720b²x²+405bx+243

and we've also got the coefficient of x³ as –720

90b³=–720===> b³=–8===> b=–2

The layout of the fast-food restaurant of dimensions 6 by 8 5 feet squares is presented in the attached table created with Sheets.

The dimensions of the facility are;

Horizontal = 6 squares

Vertical = 8 squares

The side length of each square = 5 feet

Therefore;

Area of each square = 5² ft.² = 25 ft.²

Number of squares, n, required by each dependent are therefore;

CB department, n = 300 ÷ 25 = 12 squares

CF department, n = 200 ÷ 25 = 8 squares

PS department, n = 8 squares

DD department, n = 8 squares

CS department, n = 12 squares

A layout for the fast-food restaurant is therefore;

Please see the attached table layout created using Sheets.

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__________________________________________________________

Hello! In this question, we were given an expression and would have to express it without the use of exponents.

__________________________________________________________

Explanation:

In the question, we were given:

(3a³)²

In order to express the expression without the exponents, we would have to first simplify it and then expand the expression.

__________________________________________________________

Solve:

[tex](3a^3)^2[/tex]

Simplify by "squaring" the variable inside the parenthesis, in which in this case we would multiply the "2" to the "3".

[tex]3a^6[/tex]

Now in this step, we would have to expand the expression in a different form to get rid of the exponent.

To get rid of the exponent, we would expand the " [tex]a^6[/tex] " term.

Note that the term [tex]a^6[/tex] is the same thing as a · a · a · a · a · a, it is just the expanded version of the exponent.

This means that our expressed expression without exponents, including our constant "3" would be:

3 · a · a · a · a · a · a

__________________________________________________________

Answer:

3 · a · a · a · a · a · a

__________________________________________________________

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Answer:

2x – 4y = 6

x3 – y = –2

Step-by-step explanation:

When graphing these two are linear.

Answer:

2x - 4y = –2

x3 – y = –2

Step-by-step explanation:

'cause they both have two unknown digits and this can be worked out with 3 methods;

1. substitution method

2. elimination method

3. grahical method

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the standard form equation is y=ax²+bx+c where a is the quadratic coefficient, b is the linear coefficient, and c is the constant coefficient.

therefore, for y =  5/16x² - 5x/4 -35/4

Quadratic coefficient = 5/16, linear coefficient = 5/4, constant term = -35/4

What are zeros ?

zeros denotes the factors of the given equation in other words the zeros of the function are the values that make the factors zero. The factors need to multiply out to give the original standard-form equation.

Polynomial roots are the same as polynomial zeros, so they can be found by factoring the quadratic equation into two linear factors, after which they can be equating to zero.

Its easiest to first start out with a vertex form equation because it can then be converted to a standard quadratic equation.

given a vertex at (2,-10) , and point of intersection at (6,-5), the equation can be set up like this in the form of :

y = a(x-h)²+ k

y = a(x-2)^2-10, as we know h and k from the vertex.

We also know y and x from the given point of intersection so a can be solved by substituting the values of x and y to get value of a.

y = a(x-2)²-10

-5 = a(6-2)² - 10

16a = 10-5

a = 5/16

a = 5/16 which is also known as the quadratic coefficient because it is part of a second degree quantity and a Trinomial as a whole(quadratic).

since all the variables are known, you can expand the equation and set it to standard form :

y = 5/16(x-2)²-10

y = 5/16(x-2)² - 10

y = 5/16(x²+4-4x) -10

y = 5/16x² + 5/4 - 5x/4 - 10

y =  5/16x² - 5x/4 -35/4

For reference, the standard form equation is y=ax²+bx+c where a is the quadratic coefficient, b is the linear coefficient, and c is the constant coefficient.

In this instance, 5/16x² - 5x/4 -35/4 = ax²+bx+c

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Using the Empirical Rule, it is found that 16% of men over 20 have upper arm lengths greater than 44.2 cm.

It states that, for a normally distributed random variable:

For the distribution in this problem, we have that:

Hence, 44.2 is one standard deviation above the mean, as 44.2 = 39.1 + 1 x 5.1.

The normal distribution is symmetric, meaning that 50% of the measures are above the mean and 50% are below. Of those measures above the mean, 68% are less than 44.2 and 32% are more than 44.2 cm, hence:

0.5 x 32 = 16%.

16% of men over 20 have upper arm lengths greater than 44.2 cm.

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Answer:

n(AuB)=15

n(AnB)=8 is the answer!

If Wilson produces only one 1 bag, then his profit starts decreasing

Given the function;

Let  f ( x ) = [tex]2x^4 - 4x^3 + 7[/tex]

Where;

x is  the number of bags of cereal produced

[tex]f'x = 2x^4 - 4x^3 + 7[/tex]

[tex]f'(x) = 8x^3 - 12x^2[/tex]

Factorize the expression

[tex]f'(x) = 4x^2(2x - 3)[/tex]

If the value of x is decreasing, we have;

f ′ ( x ) ≤ 0

4x² (2x - 3) ≤ 0

If x² ≥ 0, then;

2x - 3  ≤ 0

x ≤ 3/ 2

With x as the number of bags

x ∈ N

x = 1 is the only possibility

Thus, if Wilson produces only one 1 bag, then his profit starts decreasing

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The complete question:

Wilson hires a financial analyst to analyse costs and profits for his cereal production business. The analyst determines that Wilson’s eventual profit function is given as: pi = 2x ^ 4 - 4x ^ 3 + 7,  where x is the number of bags of cereal produced. At what point or number of bags of cereal will Wilson’s profit start decreasing

In the given situation (A) the median is a good measure regardless of whether the outlier is included.

To find the median:

Therefore, in the given situation (A) the median is a good measure regardless of whether the outlier is included.

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The correct question is given below:

The following dataset represents the dollar amounts of donations collected at the entrance to a free museum for one hour.

(Refer to the table given below)

Is the median a reasonably good measure of central tendency for this dataset? What if the outlier were removed from consideration?

Select the correct answer below:

(A) The median is a good measure regardless of whether the outlier is included.

(B) The median is a very poor measure regardless of whether the outlier is included.

(C) The median is a good measure when the outlier is included, but it would be a very poor measure if the outlier were removed from consideration.

(D) The median is a very poor measure when the outlier is included, but it would be a good measure if the outlier were removed from consideration.

Answer:

16

Step-by-step explanation: This is because when we look at the square we realize by definition it is a 4-sided shape with 4 equal side lengths this means that if one side is 4 all the 4 sides of a square are also 4 and in this case, we would do length times width and get an answer of an area of 16 units.

Answer:

B

Step-by-step explanation:

using the rule of exponents

[tex]a^{m}[/tex] ÷ [tex]a^{n}[/tex] = [tex]a^{(m-n)}[/tex] , then

[tex]6^{9}[/tex] ÷ 6³ = [tex]6^{(9-3)}[/tex] = [tex]6^{6}[/tex]

Answer: 31 Minutes

Step-by-step explanation:

Minus 0.80 from 7.93, and then divide 7.13 by 0.23 and you get 31

The expression sec x csc x cot x  in terms of sin x and cos x is

1/ (sin x)^2.

According to the given question.

We have an expression sec x csc x cot x.

Here, we have to write the above expression in terms of sinx and cosx.

As we know that,

sec x = 1/cos x

csc x = 1/sin x

cot x = cos x/ sin x

Thereofore, the expression sec x csc x cot x  in terms of sin x and cos x is given by

sec x csc x cot x

= (1/cos x) (1/sinx) (cosx/ sinx)

= 1/ (sin x)^2

Hence, the expression sec x csc x cot x  in terms of sin x and cos x is

1/ (sin x)^2.

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[tex] \sqrt{12.25 } = 3.5[/tex]

Taking into account the definition of a system of linear equations, 178 children, 62 students and 89 adults attended to the movie theater.

A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.

Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values ​​of the unknowns must be found, with which when are replaced in the equations, they must give the solution proposed.

In this case, a system of linear equations must be proposed taking into account that:

On the other hand, you know:

So, the system of equations to be solved consists of the following equations:

Equation 1: C + S + A= 329

Equation 2: 5C + 7S + 12A= 2392

Equation 3: A= C÷2= 1/2C

There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.

Replacing equation 3 in equation 1 and isolating the variable S you get:

C + S + 1/2C= 329

3/2C + S= 329

S= 329 - 3/2C

Replacing this expression and equation 3 in equation 2 you get:

5C + 7(329 - 3/2C) + 12×1/2C= 2392

Solving:

5C + 7×329 - 7×3/2C + 12×1/2C= 2392

5C + 2303 - 21/2C + 6C= 2392

5C - 21/2C + 6C= 2392 -2303

1/2C= 89

C= 89÷ 1/2

C= 178

Remembering that S= 329 - 3/2C, you get S= 329 - 3/2×178 → S= 62

Finally, substituting the value of C in Equation 3: A= 1/2C= 1/2×178 → A=89

In summary, 178 children, 62 students and 89 adults attended to the movie theater.

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Based on the results of the physical fitness test, the reason why we are unable to calculate the mean number of pull-us done by the students in each class is that we don't have the exact number of pull-ups done by each student.

To calculate the simple mean of a distribution of numbers, we would need the value of all the numbers to be known. These numbers will then be summed up and divided by the number of the distribution.

In this case of the physical fitness test, the number of pull-ups is not known for all the students as there are some that either did below 5 or above 10 and we don't know the exact number of pull-ups done.

As a result, we are unable to calculate the mean.

Question is:
Explain why we are unable to calculate the mean number of pull-ups done by the students in each class.

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Here we go ~

[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

[tex]\qquad \sf  \dashrightarrow \: (1 - \tan {}^{4} (a) ) \cos {}^{4} (a) [/tex]

[tex]\qquad \sf  \dashrightarrow \: (1 + \tan {}^{2} (a) )(1 - { \tan}^{2} (a)) \cos {}^{4}(a ) [/tex]

[ a² - b² = (a + b)(a - b) ]

[tex]\qquad \sf  \dashrightarrow \: ( \sec {}^{2} (a) )(1 - ( \sec{}^{2} (a) - 1) )\cos {}^{4} (a) [/tex]

[ sec² a = 1 + tan² a, so : tan² a = sec²a - 1 ]

[tex]\qquad \sf  \dashrightarrow \: \bigg( \dfrac{1}{{}cos^{2} (a)} \bigg)(2 - \sec{}^{2} (a) ) \cos {}^{4} (a) [/tex]

[tex]\qquad \sf  \dashrightarrow \: \bigg(2 - \dfrac{1}{ \cos {}^{2} (a) } \bigg) \cos {}^{2} (a) [/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{2 \cos {}^{2} (a) - 1}{ \cos {}^{2} (a) } \times \cos {}^{2} (a) [/tex]

[tex]\qquad \sf  \dashrightarrow \: 2 \cos {}^{2} (a) - 1[/tex]

[tex]\qquad \sf  \dashrightarrow \: 2(1 - \sin {}^{2} (a) ) - 1[/tex]

[ sin²a + cos² a = 1, hence sin²a = 1 - cos²a ]

[tex]\qquad \sf  \dashrightarrow \: 2 - 2 \sin {}^{2} (a) - 1[/tex]

[tex]\qquad \sf  \dashrightarrow \: 1 - 2 \sin {}^{2} (a) [/tex]

Answer:

See proof below

Step-by-step explanation:

Prove [tex]\left(\:1\:-\:tan^4A\right)cos^4A\:=\:1\:-\:2\:sin^2A[/tex]

[tex]\left(1-\tan ^4\left(A\right)\right)\cos ^4\left(A\right)[/tex]  can be expressed in sin, cos terms

Use the trigonometric identity [tex]\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}[/tex]

[tex]\left(1-\tan ^4\left(A\right)\right)\cos ^4\left(A\right) = \left(1-\left(\frac{\sin \left(A\right)}{\cos \left(A\right)}\right)^4\right)\cos ^4\left(A\right)[/tex]

[tex]\mathrm{Simplify}\:\left(1-\left(\frac{\sin \left(A\right)}{\cos \left(A\right)}\right)^4\right)\cos ^4\left(A\right)[/tex]

[tex]\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}[/tex]

= [tex]\left(1-\frac{\sin ^4\left(A\right)}{\cos ^4\left(A\right)}\right)\cos ^4\left(A\right)[/tex]

Multiplying the expression in parentheses by [tex]\cos ^4\left(A\right)[/tex] we get

[tex]\frac{\cos ^4\left(A\right)-\sin ^4\left(A\right)}{\cos ^4\left(A\right)}\cos ^4\left(A\right)[/tex]

Cancel the common factor [tex]\cos ^4\left(A\right)[/tex]

This gives us

[tex]\cos ^4\left(A\right)-\sin ^4\left(A\right)[/tex]

Now,

[tex]\sin ^4\left(A\right)=\left(\sin ^2\left(A\right)\right)^2[/tex]

[tex]\cos ^4\left(A\right)=\left(\cos ^2\left(A\right)\right)^2[/tex]

[tex]\:\cos ^4\left(A\right)-\sin ^4\left(A\right)[/tex] = [tex]\left(\cos ^2\left(A\right)\right)^2-\left(\sin ^2\left(A\right)\right)^2[/tex]

[tex]\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)[/tex]

[tex]\left(\cos ^2\left(A\right)\right)^2-\left(\sin ^2\left(A\right)\right)^2=\left(\cos ^2\left(A\right)+\sin ^2\left(A\right)\right)\left(\cos ^2\left(A\right)-\sin ^2\left(A\right)\right)[/tex]

= [tex]\cos ^2\left(A\right)-\sin ^2\left(A\right)[/tex]   [tex]\textrm{ since }\cos ^2\left(A\right)+\sin ^2\left(A\right)[/tex] = 1

Using the fact that [tex]\cos ^2\left(A\right)=1-\sin ^2\left(A\right)[/tex]

we get

[tex]\cos ^2\left(A\right)-\sin ^2\left(A\right) = 1-\sin ^2\left(A\right)-\sin ^2\left(A\right)\\\\= 1-2\sin ^2\left(A\right)[/tex]

Proved