The seventh grade class supplied bags of snacks and beverages for the school dance. They supplied 50 more beverages than bags of snacks. The dance was supplied with a total of 400 items. How many of each were supplied.
a) Define the variable(s)
b) Write a system of equations for the situation.

Answer:

y = -5

Step-by-step explanation:

-3(y+2) = 2(y+6)+ seven

-3y-6 = 2y+12+

-3y-6 = 2y + 19

-2 -2

-5y - 6 = 19

  +6 +6

-5y 25

-5 -5

 y = -5

The probability of having a 2 on first roll, 3 on second roll and 4 on third roll is 1 / 216

A fair dice is a dice that is equally likely to land on any of its faces, meaning that each face has the same probability of facing up after a roll. The probability of getting a specific face on a fair dice is represented by the number of ways that face can appear divided by the total number of possible outcomes.

For a standard six-sided fair dice, the total number of possible outcomes is 6, since there are six faces on the dice. The probability of getting a specific face (for example, the number 4) is 1/6, because there is only one way to get that number out of the six possible outcomes.

The probability of landing 2 on first roll, 3 on second roll and 4 on third roll is given as;

P = (1/6 * 1/6 * 1/6)

P = 1 / 216

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1) The bowler's scores are shown on the horizontal axis, while the frequency is shown on the vertical axis.

2) Groups of 20 points are assigned to the scoring.

3) There are 30 bowlers competing.

4) The distribution is skewed towards the left.

A histogram is a bar graph that shows data that has been divided into equal intervals in terms of frequency. The bars must touch but not overlap, and they must be of equal width.

1) A histogram shows the type of data being measured on the horizontal axis and the number of observations in each bin on the vertical axis.
Here, the horizontal axis displays the bowlers' scores, and the vertical axis displays the frequency at each interval.

2) The categories or bins of the data are listed on the horizontal axis of a histogram. The height of the columns, which represents the frequency or quantity of the occurrences, is listed on the vertical axis.
Here, the scores are arranged into 20-point groups.

3) From the histogram, it is interpreted that:

Bowlers = 2 + 4 + 6 + 8 + 9 + 1 = 30

Therefore, 30 bowlers compete

4) The distribution is skewed left. A distribution is said to be "skewed left" if the tail is to the left. The procedure of determining a "typical value" for the distribution is philosophically more challenging when the distribution is skewed.

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The required number of ways in which we can create a committee with 2 women and 2 men is 60.

Given:

We have 4 women and 5 men.

The required number of ways in which we can create a committee with 2 women and 2 men can be calculated as:

= 4C2 × 5C2

= 4!/(2!)(2!) × 5!/(2!)(3!)

= 6 × 10

= 60

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Scientific notation 295000 is 295×10³ and standard notation of  7.3×10⁻⁴ is 0.00073 and From least to greatest is 1, 3/2, π/2, √3, 9 and E is the best estimate of √3

A number system is defined as a system of writing to express numbers.

Scientific notation is a way of expressing numbers that are too large or too small to be conveniently written in decimal form.

295000 is 295×10³

Now let us write 7.3×10⁻⁴ is standard notation

7.3/10000=0.00073

We have to arrange from least to greatest

1, 9, 3/2, √3, π/2

1, 9, 1.5, 1.732, 1.57

From least to greatest is 1, 3/2, π/2, √3, 9

E is the best estimate of √3

Hence, scientific notation 295000 is 295×10³ and standard notation of  7.3×10⁻⁴ is 0.00073 and From least to greatest is 1, 3/2, π/2, √3, 9

E is the best estimate of √3

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Answer:

scalene

Step-by-step explanation:

It is a scalene triangle, meaning all sides are different lengths.  It is NOT a right triangle because the 6 cm side is "too short" to be a hypotenuse.  

Answer:

Given m∠B = 95°, a = 22 feet, and c = 17 feet, we can use the Law of Cosines to find the value of b and then use Heron's formula to find the area of ΔABC.

The Law of Cosines states:

c^2 = a^2 + b^2 - 2ab * cos(C)

Since ∠B = 95° , angle C = 180 - 95 = 85

Therefore,

c^2 = a^2 + b^2 - 2ab * cos(85)

so,

b = sqrt(c^2 - a^2 + 2ab * cos(85))

We can substitute the given values of a and c to find b:

b = sqrt(17^2 - 22^2 + 2 * 22 * 17 * cos(85))

Once we have the value of b, we can use Heron's formula to find the area of the triangle:

Area = sqrt(p*(p-a)(p-b)(p-c))

where p is the semiperimeter of the triangle: p = (a + b + c) / 2

Area = sqrt(((22 + b + 17) / 2)((22 + b + 17) / 2 - 22)((22 + b + 17) / 2 - b)*((22 + b + 17) / 2 - 17))

The only option that matches this area is 128.54 feet^2, so the area of ΔABC is 128.54 feet^2.

Answer:

Your three angle measurements are 45, 45, and 90

Step-by-step explanation:

Since we know that all the measures of a triangle add up to 180, we have:

45 + b + c = 180

Then, since it is a right triangle, we know one of the sides is 90 degrees, giving us:

45 + 90 + c = 180

Then subtract 135 from both sides:

c = 45

So, your three angle measurements are 45, 45, and 90.

Hope this helped!

The conditions for constructing a t-confidence interval are met.

In order to construct a t-confidence interval for the mean amount of time the students spend completing the homework assignment, the following conditions must be met:

The data must be a random sample from the population of interest
The sample size must be sufficiently large (typically, n > 30)
The data should be approximately normally distributed
The population standard deviation or variance must be unknown
From the information provided it can be inferred that the data is a random sample (30 students out of 75 students) and that the population standard deviation or variance is unknown. If the sample size is greater than 30, and the data is approximately normal or the sample size is large, then the conditions for constructing a t-confidence interval are met.

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The elephant weighs 8.9 × 10³ pounds more than the giraffe.

Weight is the gravitational pull of a large second object, like the Moon or the Earth, on a first object. Weight is a result of the universal law of gravitation, according to which any two objects will gravitationally attract one another with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Because of this, larger objects naturally weigh more when placed in the same location; however, the farther an object is from the Earth, the less weight it has.

Given that

Elephant weighs 1.1 × 10⁴ pounds

Giraffe weighs 2.2 × 10³ pounds

To find how much more elephant weighs than giraffe, we need to find the difference of their weights. i.e

1.1 × 10⁴ - 2.1 × 10³

= 11000 - 2100

= 8900

= 8.9 × 10³ pounds

Thus, the elephant weighs 8.9 × 10³ pounds more than the giraffe.

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Answer:

y=11.547004; x= 23.094008.

Step-by-step explanation:

Since 20 = y*sqrt*3; y=11.547004; x= 23.094008.

Answer:

5 notepads

Step-by-step explanation:

We know

Kristin spent $25 on a magazine and some notepads

1 magazine = $5

1 notepad = $4

$25 - $5 = $20

How many notepads did she buy?

We take

$20 divided by $4 = 5 notepads

So, she bought 5 notepads

Answer:

5 notepads

Step-by-step explanation:

25=4x+5

subract 5 from 25 and cancel out old 5

20=4x

divide by 4

20/4 = 5

Step-by-step explanation:

Let x be the number of bricks Ram laid.

Shyam laid x/2 bricks.

Therefore, x + x/2 + 250 = x(3/2) + 250 = x(3/2) = 300

x = (2*300)/3 = 200

Ram laid 200 bricks.

Shyam laid 200/2 = 100 bricks.

just an addition to the decent reply above

[tex]\begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we will be using this rule}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]f(x)=\log_a(x-h) \\\\[-0.35em] ~\dotfill\\\\ 1=\log_a(\frac{3}{2}-h)\implies a^1=a^{\log_a(\frac{3}{2}-h)}\implies a^1=\cfrac{3}{2}-h\implies a=\cfrac{3}{2}-h \\\\[-0.35em] ~\dotfill\\\\ -1=\log_a(3-h)\implies a^{-1}=a^{\log_a(3-h)}\implies a^{-1}=3-h\implies \cfrac{1}{a}=3-h \\\\\\ 1=(3-h)a\implies 1=(3-h)\left( \cfrac{3}{2}-h \right)\implies 1=(3-h)\left( \cfrac{3-2h}{2} \right) \\\\\\ 2=(3-h)(3-2h)\implies 2=9-9h+2h^2\implies 0=2h^2-9h+7[/tex]

[tex]0=(h-1)(2h-7)\implies \boxed{h= \begin{cases} 1\\\\ \frac{7}{2} \end{cases}}\hspace{5em}a=\cfrac{3}{2}-h\implies \boxed{a= \begin{cases} \frac{1}{2} ~~ ~~ \checkmark\\\\ -2 ~~ \bigotimes \end{cases}} \\\\\\ ~\hfill {\Large \begin{array}{llll} f(x)=\log_{\frac{1}{2}}(x-1) \end{array}} ~\hfill[/tex]

now, why we didn't use the negative value for the base "a"?

from the standpoint of a negative value raised to some exponent, is perfectly fine, however if we plug that in the change of base rule, we run into a really hot pickle.

Answer:

[tex]f(x)=\log_{\frac{1}{2}}(x-1)[/tex]

Step-by-step explanation:

Use the given points to assist in determining the logarithmic function of the given graph in the form:

Substitute the given points (³/₂, 1) and (3, -1) into the formula to create two equations:

[tex]\log_a\left(\dfrac{3}{2}-h\right)=1[/tex]

[tex]\log_a\left(3-h\right)=-1[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Low law}\\\\$\log_ab=c \iff a^c=b$\\ \end{minipage}}[/tex]

Apply the log law and rearrange each equation to isolate a:

Equation 1

[tex]\begin{aligned}\log_a\left(\dfrac{3}{2}-h\right)&=1\\\\\implies a^1&=\dfrac{3}{2}-h\\\\ a&= \dfrac{3}{2}-h\end{aligned}[/tex]

Equation 2

[tex]\begin{aligned}\log_a\left(3-h\right)&=-1\\\\\implies a^{-1}&=3-h\\\\ \dfrac{1}{a}&=3-h\\\\a&=\dfrac{1}{3-h}\end{aligned}[/tex]

Substitute the first equation into the second to eliminate a:

[tex]\dfrac{3}{2}-h=\dfrac{1}{3-h}[/tex]

Solve for h:

[tex]\implies \dfrac{3}{2}-h=\dfrac{1}{3-h}[/tex]

[tex]\implies \dfrac{3-2h}{2}=\dfrac{1}{3-h}[/tex]

[tex]\implies (3-h)(3-2h)=2[/tex]

[tex]\implies 9-9h+2h^2=2[/tex]

[tex]\implies 2h^2-9h+7=0[/tex]

[tex]\implies 2h^2-2h-7h+7=0[/tex]

[tex]\implies 2h(h-1)-7(h-1)=0[/tex]

[tex]\implies (2h-7)(h-1)=0[/tex]

[tex]\implies h=\dfrac{7}{2},\;1[/tex]

Substitute both values of h into the equations for a:

[tex]h=\dfrac{7}{2}\implies a=\dfrac{3}{2}-\dfrac{7}{2}=-2[/tex]

[tex]h=\dfrac{7}{2}\implies a=\dfrac{1}{3-\dfrac{7}{2}}=-2[/tex]

[tex]h=1\implies a=\dfrac{3}{2}-1=\dfrac{1}{2}[/tex]

[tex]h=1\implies a=\dfrac{1}{3-1}=\dfrac{1}{2}[/tex]

Therefore, the two values of h given is two possible values of a. Since a is the base of the function, and the bases of logarithmic function cannot be negative, the value of a cannot be -2.  Therefore, the only valid value of a is ¹/₂.

Similarly, logs of negative numbers are undefined, therefore the value of h cannot be ⁷/₂ since this would make the argument negative for the points given.  Therefore, the only valid value of h is 1.

Inputting the found values of a and h into the given formula, the logarithmic function of the given graph is:

Answer:

You would get your first ace at card 10.6 or about card 11.

600 ≤ X < 950 are the possible values of the number of dollars in the original pile of money.

The difference between the highest and lowest values in statistics for a particular data collection is called the range. As an illustration, if the data set contains 1, 5, 8, 12, 3, the range will be 12 - 1 = 11. The difference between the greatest and lowest observation might thus also be used to determine the range.

We need to first put the data in ascending order before we can determine the range of a particular set of observations. To determine the range, calculate the difference between the highest and minimum values afterwards.

Let X be the amount in the pile.

What is left after putting $200 in the pile = X - $200

4/5 = 0.8

What she gives away = 0.8 (X - 200)

The remaining is : 0.2 (X - 200) what she puts in her right pocket.

200 + 0.2(X - 200) > X - 600

200 - 0.2X - 40 > X - 600

Solving the inequality we have :

= 160 + 0.2x > X - 600

= 760 > X - 0.2X

= 760 > 0.8X

= 950 > X

This implies : X < 950

It is said : X ≥ 600

The possible range of X is thus : 600 ≤ X < 950

thus, 600 ≤ X < 950 are the possible values of the number of dollars in the original pile of money.

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The value of x 1/2 or 3/2.

The logarithm of any number N if interpreted as an exponential form, is the exponent to which the base of the logarithm should be raised, to obtain the number N. Here we shall aim at knowing more about logarithmic functions, types of logarithms, the graph of the logarithmic function, and the properties of logarithms.

The basic logarithmic function is of the form f(x) = logax (r) y = logax, where a > 0. It is the inverse of the exponential function ay = x. Log functions include natural logarithm (ln) or common logarithm (log). Logarithmic function properties are helpful to work across complex log functions. All the general arithmetic operations across numbers are transformed into a different set of operations within logarithms. The product of two numbers, when taken within the logarithmic functions is equal to the sum of the logarithmic values of the two functions. Similarly, the operations of division are transformed into the difference of the logarithms of the two numbers.

Here, we use:

1) loga/b = log a - log b

2) logba = (logc a)/(logc b)

Given logarithmic equation:

logₓ(8x-3) - logₓ4 = 2

logₓ (8x-3/4) = 2

(8x-3/4) = x²

8x - 3 = 4x²

4x² - 8x + 3 = 0

x = 1/2 or 3/2

Thus, the value of x 1/2 or 3/2.

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Answer:

present balance=782

Step-by-step explanation:

65+756=821

now has 821 in account

821-44.10=776.9

now has 776.9

776.9-4.90=772

now has 772 in account

772+10=782

he now has 782 in his account

The volume of the sphere with a great circle of radius 12 cm is 7238.2 cm³. The correct option is A) 7,238.2 cm3

From the question, we are to calculate the volume of the sphere with the given radius.

From the formula for calculating the volume of a sphere, we have that

V = 4/3πr³

Where V is the volume

and r is the radius

From the given information,

The sphere has a great circle of radius 12 cm

Thus,

r = 12 cm

Substitute the value of r into the equation,

V = 4/3πr³

That is,

V = 4/3 × π × 12³

V = 4/3 × π × 1728

V = 4 × π × 576

V = 4 × 576 × π

V = 2304π

V = 7238.2 cm³

Hence, the volume is 7238.2 cm³

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Answer:

Say you take a pizza and cut it perfectly down the middle so there are two slices. Each slice is a half or 1/2 of the pizza.

A scatterplot that shows a negative linear association.

A scatterplot that is most likely to have a line of best fit represented by the equation y = -5x + 2 is the one where the data points have a strong negative linear association. This means that as the x-values increase, the y-values decrease. The slope of the line is -5, which is negative, indicating a negative association. The y-intercept of the line is 2, which represents the point where the line crosses the y-axis.

A scatterplot that shows a negative linear association would have its points distributed in a downward direction from left to right, with a slope of -5.

Therefore, A scatterplot shows a negative linear association will most likely have a line of best fit

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The inequality 60 + 5x ≤ 130 shows the maximum width, in inches, of the box that meets the restrictions of the shipping service, given that the box has a height of 60 inches and a length of 2.5 times the width.

1. Let x = the width of the box, in inches.

2. The perimeter of the base of the box is calculated by the following formula: Perimeter = 2(Width) + 2(Length)

3. Therefore, the equation is: 60 + 5x + 2(2.5x) = 130

4. Simplify the equation by expanding the parentheses and combining like terms: 60 + 5x + 5x = 130

5. Solve for x: 5x = 130 - 60

6. Simplify: 5x = 70

7. Divide both sides by 5 to find x: x = 70/5

8. Simplify: x = 14

9. Therefore, the inequality 60 + 5x ≤ 130 shows the maximum width, in inches, of the box that meets the restrictions of the shipping service is 14 inches

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To solve the differential equation y" + y = csc(x) using variation of parameters, we first need to find a particular solution.

Since the non-homogeneous term is csc(x), we can try a particular solution of the form yp = u(x) csc(x), where u(x) is a function to be determined.

Substituting yp into the differential equation:

u''(x)csc(x) + u(x)csc(x) = csc(x)

To find u(x), we can multiply both sides by csc(x) and integrate with respect to x:

∫u''(x)csc(x) dx + ∫u(x)csc^2(x) dx = ∫csc(x) dx

Integrating the first term by parts:

u'(x) = ∫csc(x) dx = ln|cot(x/2)| + C1

Integrating the second term by parts:

u(x) = -∫csc(x)cot(x) dx = -ln|sin(x)| + C2

So, the particular solution for yp is:

yp = u(x)csc(x) = -ln|sin(x)|csc(x) + C1csc(x) + C2

Now we need to find the general solution y = yc + yp

where yc is the general solution for the complementary homogeneous equation y" + y = 0

yc = C1cos(x) + C2sin(x)

Now we can use the method of variation of parameters to find the general solution for the non-homogeneous equation

By this method, we need to find two functions v1(x) and v2(x) such that W(yc, yp) = v1(x)cos(x) + v2(x)sin(x)

W(yc, yp) = (yc * yp') - (yp * yc')

W(yc, yp) = (C1cos(x) + C2sin(x)) * (-cot(x)csc(x) + C1) - (-ln|sin(x)|csc(x) + C1csc(x) + C2) * (-sin(x)cos(x) + cos(x)sin(x))

W(yc, yp) = -C1cot(x)csc(x) + C1^2 + C2csc(x)

As W(yc, yp) is not identically zero, we can proceed to find v1(x) and v2(x)

v1'(x) = -cot(x)csc(x) + C1

v2'(x) = csc(x)

Integrating both sides by parts:

v1(x) = ln|cot(x/2)| - cot(x) + C3

v2(x) = -cot(x) + C4

The general solution for the non-homogeneous equation y" + y = csc(x) is:

y = yc + yp = (C1cos(x) + C2sin(x)) + (sin(x)ln|sin(x)|

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The amount of vanilla extract that will be needed to be able to make 2 dozen cookies is 2 teaspoons of vanilla extract

The recipe given will be able to yield 1 . 5 dozen of Oatmeal Raisin Cookies  which means that the 1 . 5 teaspoons of vanilla extract will be needed for 1. 5 dozens of Oatmeal Raisin Cookies.

If you are to make 2 dozen cookies therefore, it means that the recipe would require a higher amount of vanilla extract which can be found by the formula:

= ( Dozen cookies needed x teaspoons in  1 .5 dozen ) / 1 . 5 dozen

= ( 2 x 1 . 5 ) / 1 . 5

= 2 teaspoons of vanilla extract

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The height of the tree to the nearest hundredth of a meter is 29.1m.

We can use the concept of similar triangles to find the height of the tree. The ratio of the height of the tree to the length of its shadow is the same as the ratio of Hunter's height to the length of his shadow.

Let h be the height of the tree.

Hunter's height : h = 1.75m : x (where x is the height of the tree)

Hunter's shadow : Tree's shadow = 1.75m : 30.85m

We know that Hunter is 26.8 meters away from the tree, and the tip of his shadow meets the tip of the tree's shadow.

Hunter's shadow = Hunter's height + Tree's shadow

1.75 + x = 30.85

x = 29.1m

Therefore, the height of the tree to the nearest hundredth of a meter is 29.1m.

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The solution for the variable of each expression is given as follows:

D) 3 - 2(s - 1) = 13 + 6s: s = -1.

E) 2(n + 9) = -6(2n - 5) + 8: n = 10/7.

F) 5(4k - 3) - 5k = 10 + 2(3k + 1): k = 3.

The expressions are solved applying the distributive property when needed, then combining the like terms, and finally isolating the variable.

The expression for item d is given as follows:

3 - 2(s - 1) = 13 + 6s

Hence:

3 - 2s + 2 = 13 + 6s -> Distributive property.

8s = 5 - 13 -> Combine the like terms.

8s = -8

s = -8/8

s = -1.

The expression for item e is given as follows:

2(n + 9) = -6(2n - 5) + 8

2n + 18 = -12n + 30 + 8 -> Distributive property.

14n = 20 -> Combine the like terms.

n = 20/14

n = 10/7.

The expression for item f is given as follows:

5(4k - 3) - 5k = 10 + 2(3k + 1)

20k - 15 - 5k = 10 + 6k + 2

15k - 15 = 12 + 6k

9k = 27

k = 27/9

k = 3.

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The 205th term of the given geometric sequence is given by 6

A geometric sequence is a special type of sequence where the ratio of every two successive terms is a constant. This ratio is known as a common ratio of the geometric sequence.

Given here: The first term of a given geometric sequence as 6 and second term -6.

Thus clearly we can deduce that the common ratio of the sequence is given by -1

Therefore the geometric sequence has two elements that are alternating sequence of the form 6,-6,6,-6,6.........

Therefore the 205th term of the sequence will be given by

T₂₀₅=6×-1²⁰⁵⁻¹

       =6×1

        =6

Hence, The 205th term of the given geometric sequence is given by 6

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The solution of the equation is as follows;

Equations are mathematical statements containing two algebraic expressions on both sides of an 'equal to (=)' sign.

The equation can be solved as follows:

x - 10 = √9x

subtract x from both sides of the equation

x - x - 10  = √9x - x

-10 =  √9 x - x

-10 = 3x - x

-10 = 2x

divide both sides by 2

x = -10 / 2

x = -5

Therefore,

x = -5

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