At an aquarium, 6 out of 18 deliveries are plants. Out of 15 deliveries in one week,how many are plants?

Answers

Answer 1
The answer to your question is:

5 deliveries.

Related Questions

Omar's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Omar $4.85 per pound, and type B coffee costs $5.95 per pound. This month's blend used twice as many pounds of type B coffee as type A, for a total cost of $519.25 . How many pounds of type A coffee were used?

Answers

27 pounds of Type A coffee 

x = 2, y = -1
14
2. x = 0, y = 2.5
1.665
3. x = -1, y = -3
0.44
4. x = 0.5, y =
9.17
5. x = , y =
-1
6. x = √2, y = √2
-11.25

Answers

Answer:

b

Step-by-step explanation:

The area of a rectangle wall of a barn is 216 ft.² it's length is 6 feet longer than twice it's width. find the length and width of the wall of the barn

Answers

the answer is that the length is 24 and the width is 9 because 9•2= 18+6=24 and 24•9= 216 ft²

A map is scaled so that 3 cm on the map is equal to 21 actual miles. if two cities on the map are 5 cm apart, what proportion would you use to solve the problem?

Answers

3/21 = 5/x....3 cm to 21 miles = 5 cm to x miles

The sum of 7 consecutive odd numbers is 91. What is the sum of the two largest numbers in this set?

Answers

7 consecutive odd numbers...x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12

x + x + 2 + x + 4 + x + 6 + x + 8 + x + 10 + x + 12 = 91
7x + 42 = 91
7x = 91 - 42
7x = 49
x = 49/7
x = 7

x + 2 = 7 + 2 = 9
x + 4 = 7 + 4 = 11
x + 6 = 7 + 6 = 13
x + 8 = 7 + 8 = 15
x + 10 = 7 + 10 = 17
x + 12 = 7 + 12 = 19

the sum of the 2 largest numbers is : 17 + 19 = 36 <==

how do you write 20484163 in different forms

Answers

You mean 20,484,163?

If so..

Standard form would be: 20,484,163

Word form would be: twenty million, four hundred eighty-four thousand, one hundred sixty-three

Expanded form would be: 20,000,000 + 400,000 + 80,000 + 4,000 + 100 + 60 + 3

The brightness of a variable star adds a component to the simple harmonic motion we have studied in this lesson. Since the brightness is variable the vertical axis may no longer be equal to zero. Also included in the variance is a phase shift. In this case, the equation for this function would be: y = a cos w(t – c ) + b.

Suppose we have a variable star whose brightness alternately increases and decreases. For this star, the time between periods of maximum brightness is 6.5 days. The average brightness (or magnitude) of the star is 5.0 and its brightness varies by + 0.25 magnitude.


1. What is the amplitude of the function for this model?

2. What is the period?

3. What is w?

4. What is the vertical shift?

5. Is there a phase shift? If so, what is it?

6. What is the function

Answers

The given model is
y a cos w(t - c) + b

1. The brightness varies by  0.25. Therefore the amplitude is
   a = 0.25

2. The time between maximum brightness is 6.5 days.
    Therefore the period is T = 6.5 days

3. By definition,
    w = 2π/T. Therefore,
    w = 2π/6.25 = 0.967

4. The average brightness is 5, therefore the vertical shift is b = 5.

5. Assume that maximum brightness occurs at the time, t = 0.
    Therefore there is no phase shift so that c = 0.

6. The function is
     y = 0.25 cos(0.967t) + 5

A graph of the function is shown below.
   

61702 67102 same or different?

Answers

The numbers differ by 5,400, with 67,102 being greater than 61,702.

Let me know if you need any more help!

Solve cos x +sqr root of 2 = -cos x for x over the interval 0,2pi

Answers

[tex]\bf cos(x)+\sqrt{2}=-cos(x)\implies 2cos(x)+\sqrt{2}=0\implies 2cos(x)=-\sqrt{2} \\\\\\ cos(x)=-\cfrac{\sqrt{2}}{2}\implies \measuredangle x= \begin{cases} \frac{3\pi }{4}\\ \frac{5\pi }{4} \end{cases}[/tex]

Suppose you have a normally distributed set of data pertaining to a standardized test. the mean score is 1000 and the standard deviation is 200. what is the z-score of 900 point score? 1.5 1 0.5 â0.5

Answers

The z-score is calculated by teh equation,
                       z-score  = (X - μ) / σ
where X is the given data, μ is the mean/average and σ is the standard deviation.

From the given above,
X = 900
μ = 1000
σ = 200

Substituting the known values from the given above to the equation,
                        z-score = (900 - 1000)/ 200
                        z-score = (-100)/200
                        z-score = -1/2 = -0.5

Hence, the value of z-score is equal to -0.5. 

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section 4.2, y2 = y1(x) e−∫p(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). 9y'' − 12y' + 4y = 0; y1 = e2x/3

Answers

Given that [tex]y_1=e^{2x/3}[/tex], we can use reduction of order to find a solution [tex]y_2=v(x)y_1=ve^{2x/3}[/tex].

[tex]\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}[/tex]
[tex]\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}[/tex]

[tex]\implies9y''-12y'+4y=0[/tex]
[tex]\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0[/tex]
[tex]\implies9v''-3v'=0[/tex]

Let [tex]u=v'[/tex], so that

[tex]9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0[/tex]
[tex]e^{-x/3}u'-\dfrac13e^{-x/3}u=0[/tex]
[tex]\left(e^{-x/3}u\right)'=0[/tex]
[tex]e^{-x/3}u=C_1[/tex]
[tex]u=C_1e^{x/3}[/tex]

[tex]\implies v'=C_1e^{x/3}[/tex]
[tex]\implies v=3C_1e^{x/3}+C_2[/tex]

[tex]\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}[/tex]
[tex]\implies y_2=3C_1e^x+C_2e^{2x/3}[/tex]

Since [tex]y_1[/tex] already accounts for the [tex]e^{2x/3}[/tex] term, we end up with

[tex]y_2=e^x[/tex]

as the remaining fundamental solution to the ODE.

The indicated function y1(x) is a solution of the given differential equation.The general solution is [tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

What is a differential equation?

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.

The derivatives might be of any order, some terms might contain the product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

Given differential equation is

y''-4y'+4y=0

and

[tex]y_1(x) = e^{2x}[/tex]

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx[/tex]

The general form of equation

y''+P(x)y'+Q(x)y=0

Comparing both the equation

So, P(x)= - 4

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx\\\\\\[/tex]

[tex]y_2(x) = e^{2x}\int e^{-4x} \, / e^{4x}dx[/tex]

[tex]y_2(x) = e^{2x}\int e^{-8x}dx\\\\y_2(x) = -e^{-6x}/8[/tex]

The general solution is

[tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

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A football is punted from a height of 2.5 feet above the ground with an initial vertical velocity of 45 feet per second.
Write an equation to model the height h in feet of the ball t seconds after it has been punted.
The football is caught at 5.5 feet above the ground. How long was the football in the air?

Answers

The standard kinematics equation is for an object projected vertically is:
H(t)=H0+v0(t)+(1/2)at^2
H0=height at time 0
v0(t)=vertical velocity at time 0
a=acceleration [equals -g for gravity]
H(t) height of projectile at time t.

We're given
H0=2.5'
v0=45 '/s
a=-32.2 '/s^2

so the kinematics equation is
H(t)=2.5+45(t)+(1/2)(-32.2)t^2= 2.5+45t-16.1t^2

Solve for H(t)=5.5 using the quadratic formula:
H(t)=5.5= 2.5+45t-16.1t^2
t=0.0683 s  [ on its way up ], or
t=2.727 s [caught on its way down]

Therefore the football was in the air for 2.727 seconds.

Find the area under the standard normal distribution curve to the left of z=-2.15 and to the right of z=1.62

Answers

The area under the standard normal distribution represents probability from 0 to 1.
So, what's being asked, in essence, is what is P(Z ≤ -2.15) and P(Z ≥ 1.62).
P(Z ≤ -2.15) = 1 - P(Z ≤ 2.15) = 1 - 0.9842 = 0.0158.
P(Z ≥ 1.62) = 1 - P(Z ≤ 1.62) = 1 - 0.9474 = 0.0526

Using the normal distribution table, which shows the percentage of the areas to the left a normal distribution, the area to the left z = - 2.15 and area to the right of z = 1.62 are 0.0158 and 0.0526 respectively.

1.)

The area under the normal distribution curve to the left of z = 2.15 can be expressed thus :

P(Z ≤ -2.15)

Using a normal distribution table ; the area to the left is

P(Z ≤ -2.15) = 0.0158

2.)

The area under the normal distribution curve to the right of z = 1.62 can be expressed thus :

P(Z ≥ 1.62) = 1 - P(Z ≤ 1.62)

Using a normal distribution table ; the area to the left is P(Z ≤ 1.62) = 0.94738

P(Z ≥ 1.62) = 1 - 0.94738 = 0.0526

Therefore, the area to the left z = - 2.15 and area to the right of z = 1.62 are 0.0158 and 0.0526 respectively.

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 Donna and Phyllis are reviewing for a math final. They have 150 pages to cover. Donna can review 37 pages an hour, but Phyllis can only go 2/3 as fast as Donna. About how long does it take each girl to finish?

Answers

Donna: 37 pages per hour.  There are 150 pages.
Time: 150 pages  / (37 pages/hour)
= 150/37 hours
= 4 hours 3 1/2 min. (approx.)

Phyllis: (2/3)*37 pages per hour.  There are 150 pages.
Time: 150 pages / ((2/3)*37 pages / hour )
= 150*3/(2*37) hours
= 225/37 hours
= 6 hours 5 minutes (approx.)

Donna will finish reviewing in about 4.05 hours, while Phyllis will take approximately 6.08 hours.

Donna and Phyllis are reviewing for a math final and have 150 pages to cover. Let's calculate how long it takes each girl to finish reviewing:

Donna can review 37 pages an hour. To find the time she needs, we use the formula:

Time = Total Pages / Pages per Hour
Time = 150 / 37
Time ≈ 4.05 hours

Phyllis reviews at 2/3 of Donna's speed. So, her rate is:

Phyllis' rate = (2/3) × 37 ≈ 24.67 pages per hour

To find the time Phyllis needs, we use the same formula:

Time = Total Pages / Pages per Hour
Time = 150 / 24.67
Time ≈ 6.08 hours

In summary, Donna will finish reviewing in about 4.05 hours, while Phyllis will take approximately 6.08 hours.

Which of the following statements is not true?

An angle bisector can be a median of a triangle.
A perpendicular bisector can be an altitude of a triangle.
A median can be an altitude of a triangle.
All of the statements are true.

Answers

All of the statements has the qualifier "can be".
This means that we need just one single in each example to make the statement true.

In an equilateral triangle, medians, angle bisectors, altitudes and perpendicular bisectors are all coincident, which makes the first three statements true.  This in turn makes the fourth statement true.

So there are no false statements.

Which value is a discontinuity of x^2+7x+1/x^2+2x-15? x=-1 x=-2 x=-5 x=-4

Answers

[tex]\dfrac{x^2+7x+1}{x^2+2x-15}=\dfrac{x^2+7x+1}{(x+5)(x-3)}[/tex]

which is undefined when [tex]x=-5[/tex] or [tex]x=3[/tex]. The answer is then the third choice.
Final answer:

The value of x=-5 is a discontinuity of the function x^2+7x+1 / x^2+2x-15 since it makes the denominator of this function equal to zero.

Explanation:

The subject of this question is the discontinuity of a rational function. In Mathematics, a function f(x) = (p(x))/(q(x)), where p(x) and q(x) are polynomials, is said to be discontinuous at a particular value of x if and only if q(x) = 0 at that value. From the equation in the question; x^2+7x+1/x^2+2x-15, we can determine its discontinuity by finding the values of x that would make the denominator equal to zero. This is done by solving the polynomial equation x^2+2x-15 = 0 for x. The solutions to this equation represent the values at which the function is discontinuous.

By applying the quadratic formula, (-b ± sqrt(b^2 -4ac))/(2a), where a = 1, b = 2, and c = -15, we get that x = -5, and 3. However, the values given in the question are x=-1, x=-2, x=-5, and x=-4. From these options, only x=-5 makes the denominator zero, thus, x = -5 is a point of discontinuity in the function x^2+7x+1 / x^2+2x-15.

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If the APR of a savings account is 3.6% and interest is compounded monthly, what is the approximate APY of the account?

Answers

The Answer is ''3.66%''

Answer:

3.66% ( approx )

Step-by-step explanation:

Since, the formula of annual percentage yield is,

[tex]APY = (1+\frac{r}{n})^n-1[/tex]

Where,

r = stated annual interest rate,

n = number of compounding periods,

Here, r = 3.6% = 0.036,

n = 12 ( ∵ 1 year = 12 months )

Hence, the annual percentage yield is,

[tex]APY=(1+\frac{0.036}{12})^{12}-1=1.03659 - 1 = 0.036599\approx 0.0366 = 3.66\%[/tex]

What is the midpoint of the line segment (-3,-2) and (1,4)

Answers

Answer:

  (-1, 1)

Step-by-step explanation:

The midpoint (M) is found by averaging the coordinates of the end points:

  M = ((-3, -2) +(1, 4))/2 = ((-3+1)/2, (-2+4)/2) = (-2/2, 2/2)

  M = (-1, 1)

The midpoint of the line segment is (-1, 1).

You roll a pair of fair dice until you roll “doubles” (i.e., both dice are the same). what is the expected number, e[n], of rolls?

Answers

Final answer:

The expected number of rolls, e[n], to get doubles on a pair of fair dice is 6. This is calculated by recognizing that getting doubles has a probability of 1/6, and the expectation for a geometric distribution is the inverse of the probability (1/p).

Explanation:

To solve the problem of finding the expected number of rolls, e[n], to get doubles on a pair of fair dice, we first need to calculate the probability of rolling doubles. Since there are 6 faces on each die, there are a total of 6 x 6 = 36 possible outcomes when rolling two dice.

Out of these 36 possible rolls, there are 6 outcomes that result in doubles: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). This means the probability of getting doubles in one roll is 6/36, which simplifies to 1/6.

Because each roll is independent, we can model the scenario using a geometric distribution, where the expected value, or mean, is given by 1/p, where p is the success probability. Substituting p with 1/6, the expected number of rolls to get doubles would be 1/(1/6) = 6.

Therefore, the expected number of rolls needed to roll doubles is 6.

A cone-shaped paper drinking cup is to be made to hold 36 cm3 of water. find the height and radius of the cup that will use the smallest amount of paper. (round your answers to two decimal places.)

Answers

The formula for volume of cone is:

V = π r^2 h / 3

or

π r^2 h / 3 = 36 cm^3

Simplfying in terms of r:

r^2 = 108 / π h

To find for the smallest amount of paper that can create this cone, we call for the formula for the surface area of cone:

S = π r sqrt (h^2 + r^2)

S = π sqrt(108 / π h) * sqrt(h^2 + 108 / π h) 

S = π sqrt(108 / π h) * sqrt[(π h^3 + 108) / π h] 

Surface area = sqrt (108) * sqrt[(π h + 108 / h^2)] 

Getting the 1st derivative dS / dh then equating to 0 to get the maxima value:

dS/dh = sqrt (108) ((π – 216 / h^3) * [(π h + 108/h^2)^-1/2] 

Let dS/dh = 0 so,

 π – 216 / h^3 = 0 

h^3 = 216 / π

h = 4.10 cm

Calculating for r:

r^2 = 108 / π (4.10)

r = 2.90 cm

 

Answers:

 h = 4.10 cm

r = 2.90 cm

The height of the cone is [tex]\boxed{4.10}[/tex] and the radius of the cone is [tex]\boxed{2.90}.[/tex]

Further explanation:

The volume of the cone is [tex]\boxed{V = \dfrac{1}{3}\left( {\pi {r^2}h} \right)}.[/tex]

The surface area of the cone is [tex]\boxed{S=\pi \times r\times l}[/tex]

Here l is the slant height of the cone.

The value of the slant height can be obtained as,

[tex]\boxed{l = \sqrt {{h^2} + {r^2}} }[/tex].

Given:

The volume of the cone shaped paper drinking cup is [tex]36{\text{ c}}{{\text{m}}^3}[/tex].

Explanation:

The volume of the cone shaped paper drinking cup  [tex]36{\text{ c}}{{\text{m}}^3}[/tex].

[tex]\begin{aligned}V&=36\\\frac{1}{3}\left({\pi {r^2}h}\right) &= 36\\{r^2}&= \frac{{108}}{{\pi h}}\\\end{aligned}[/tex]

The surface area of the cone is,

[tex]\begin{aligned}S &= \pi\times\sqrt {\frac{{108}}{{\pi h}}}\times\sqrt {{h^2} + \frac{{108}}{{\pi h}}}\\&= \sqrt{108}\times\sqrt{\frac{{\pi {h^3} + 108}}{{\pi h}}}\\&=\sqrt {108}\times\sqrt {\pi h + \frac{{108}}{{{h^2}}}}\\\end{aligned}[/tex]

Differentiate above equation with respect to h.

[tex]\dfrac{{dS}}{{dh}}=\sqrt {108}\times \left( {\pi  - \dfrac{{216}}{{{h^3}}}}\right)\times {\left( {\pi h + \dfrac{{108}}{{{h^2}}}}\right)^{ - \dfrac{1}{2}}}[/tex]

Substitute 0 for [tex]\dfrac{{dS}}{{dh}}[/tex].

[tex]\begin{aligned}\pi- \dfrac{{216}}{{{h^3}}}&= 0\\\dfrac{{216}}{{{h^3}}}&= \pi\\\dfrac{{216}}{{3.14}} &= {h^3}\\h &= 4.10\\\end{aligned}[/tex]

The radius of the cone can be obtained as,

[tex]\begin{aligned}{r^2}&=\frac{{108}}{{\pi \left({4.10} \right)}}\\{r^2}&= \frac{{108}}{{3.14 \times 4.10}}\\{r^2}&= 8.40\\r&= \sqrt {8.40}\\r &= 2.90\\\end{aligned}[/tex]

Hence, the height of the cone is  [tex]\boxed{4.10}[/tex]and the radius of the cone is [tex]\boxed{2.90}[/tex].

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Mensuration

Keywords: cone shaped paper, drinking cup, volume, [tex]36{\text{ c}}{{\text{m}}^3}[/tex], height of cone, cup, smallest amount of paper, water.

13-36x^2=-12 which value of x is a solution to he equation

Answers

13-36x^2=-12

36x^2=-12-13
x^2=24/36

×1=-5/6 ×2=5/6

What standard deviation below the mean of normal young adults equals osteoporosis?

Answers

The standard deviation from the mean of a young (30-year old) adult is called a t-score.
If the t-score is -1, it means that the bone density is 1 standard deviation below the mean.
It is generally considered a t-score between -1 and -2.5 low bone density.
Patients with t-scores below -2.5 (e.g. -3) is considered suffering from osteoporosis.  Also, patients within this range AND suffered from one or more fractures is considered established osteoporosis.

Explain why rationalizing the denominator does not change the value of the original expression

Answers

Rationalizing is just simpllifying, so the simplified value has the same value as the original expression.

Answer:

Because basically, you are multiplying by 1

Step-by-step explanation:

Let me explain this with an example. Rationalize the following expression:

[tex]\frac{5}{\sqrt{7} }[/tex]

In order to rationalize the denominator, the numerator and denominator of the fraction must be multiplied by the root of the denominator. So, what happen if you do that? Well first of all you aren't altering the expression because:

[tex]\frac{\sqrt{7} }{\sqrt{7} } =1[/tex]

Right? Because a certain quantity divided by itself is always equal to 1. So basically you are doing this because you want to rewrite the expression without altering its original value, it is the same when you do this:

[tex]9=3^2=3+3+3=\sqrt{81}[/tex]

Therefore, the only thing you do when you rationalize is remove radicals from the denominator of a fraction. Take a look:

[tex]\frac{5}{\sqrt{7} } *\frac{\sqrt{7} }{\sqrt{7} } =\frac{5*\sqrt{7} }{\sqrt{7}*\sqrt{7} } =\frac{5*\sqrt{7}}{(\sqrt{7} )^2} =\frac{5*\sqrt{7} }{7}[/tex]

You can check this new expression is equal to the original using a calculator:

[tex]\frac{5}{\sqrt{7} } \approx1.8898\\\\\frac{5*\sqrt{7} }{7} \approx1.8898[/tex]

The area of one triangle is 150 square centimeters when it's height is 20 centimeters and it's base length is 15 centimeters. What is the area of a triangle having a height of 30 centimeters and a base length of 18 centimeters.

Answers

Formula for finding the area of a triangle: [tex]A = \frac{1}{2}bh[/tex]
Where b is the base and h is the height.

Substitute the variables for the values.

[tex]A = \frac{1}{2}(30)(18)[/tex]
[tex]A = \frac{1}{2}540[/tex]
A = 270

So, the area is 270 square centimeters.


The bookstore ordered 15 Sociology work books and 10 Economy 101 textbooks on August 1st to prepare for the fall semester. After realizing they miscounted, the bookstore order 1 more Sociology workbook and three more Economy 101 textbooks on August 15th. If August 1st the order was for $4300 and the August 15th order was $800 , find the cost of one Sociology workbook and one Economy 101 textbook?

Answers

e = number of books on Economy
s = number of books on Sociology

so the order on August 1st looks like 15s + 10e = 4300
and the order on August 15th looks like 1s + 3e = 800

[tex]\bf \begin{array}{lllll} 15s&+&10e&=&4300\\ s&+&3e&=&800 \end{array}\\\\ -------------------------------\\\\ \boxed{s}=800-3e\qquad thus \\\\\\ 15s+10e=4300\implies 15\left( \boxed{800-3e} \right)+10e=4300\qquad then \\\\\\ 12000-45e+10e=4300\implies 12000-4300=45e-10e \\\\\\ 7700=35e\implies \cfrac{7700}{35}=e\implies \boxed{220=e} \\\\\\ now \qquad s+3e=800\implies s+3(220)=800\implies s=800-660 \\\\\\ \boxed{s=140}[/tex]
Final answer:

The cost of one Sociology workbook is $140, and the cost of one Economy 101 textbook is $220, determined by solving two simultaneous equations based on the initial and additional orders placed by the bookstore.

Explanation:

To solve for the cost of one Sociology workbook and one Economy 101 textbook, let's label the cost of one Sociology workbook as S and the cost of one Economy 101 textbook as E. Initially, the bookstore ordered 15 Sociology workbooks and 10 Economy 101 textbooks, spending a total of $4300. This gives us the equation: 15S + 10E = 4300. Later, the store ordered an additional 1 Sociology workbook and 3 Economy 101 textbooks for $800, leading to the equation: 1S + 3E = 800.

To solve these equations simultaneously, we can multiply the second equation by 15 to get 15S + 45E = 12000 and subtract the first equation from it: (15S + 45E) - (15S + 10E) = 12000 - 4300, simplifying to 35E = 7700. Dividing both sides by 35 gives us E = 220. Substituting E = 220 back into the first equation 15S + 10(220) = 4300 gives us 15S + 2200 = 4300, thus 15S = 2100 and S = 140.

Therefore, the cost of one Sociology workbook is $140 and the cost of one Economy 101 textbook is $220.

The function for the cost of materials to make a shirt is f(x) = five sixths x + 5, where x is the number of shirts. The function for the selling price of those shirts is g(f(x)), where g(x) = 5x + 6. Find the selling price of 18 shirts

Answers

[tex]\bf \begin{cases} f(x)=\cfrac{5}{6}x+5\\\\ g(x)=5x+6 \end{cases}\qquad g(\ f(x)\ )=5[\ f(x)\ ]+6 \\\\\\ f(18)=\cfrac{5}{6}(18)+5\implies f(18)=\cfrac{5\cdot 18}{6}+5\implies f(18)=15+5 \\\\\\ \boxed{f(18)=20}\\\\ -------------------------------\\\\ g(\ f(18)\ )=5[\ f(18)\ ]+6\implies g(\ f(18)\ )=5[\ 20\ ]+6 \\\\\\ g(\ f(18)\ )=106[/tex]

What is equivalent to the expression "the quotient of five and seven"?

Answers

5/7 
five divided by seven=0.7142857143
5 divided by 7 = 0.714285714285714

The average score on a standardized test is 500 points with a standard deviation of 50 points. If 2,000 students take the test at a local school, how many students do you expect to score between 500 and 600 points?

Answers

To solve this problem, we use the z statistic. The formula for z score is given as:

z = (x – u) / s

Where,

x = sample score

u = the average score = 500

s = standard deviation = 50

 

First, we calculate for z when x = 500

z = (500 – 500) / 50

z = 0 / 50

z = 0

Using the standard z table, at z = 0, the value of P is: (P = proportion)

P (z = 0)= 0.5

 

Secondly, we calculate for z when x = 600

z = (600 – 500) / 50

z = 100 / 50

z = 2

Using the standard z table, at z = 2, the value of P is: (P = proportion)

P (z = 2) = 0.9772

 

Since we want to find the proportion between 500 and 600, therefore we subtract the two:

P (500 ≥ x ≥ 600) = 0.9772 – 0.5

P (500 ≥ x ≥ 600) = 0.4772

 

Answer:

Around 47.72% of students have score from 500 to 600.

Answer:

To solve this problem, we use the z statistic. The formula for z score is given as:

z = (x – u) / s

Where,

x = sample score

u = the average score = 500

s = standard deviation = 50

First, we calculate for z when x = 500

z = (500 – 500) / 50

z = 0 / 50

z = 0

Using the standard z table, at z = 0, the value of P is: (P = proportion)

P (z = 0)= 0.5

Secondly, we calculate for z when x = 600

z = (600 – 500) / 50

z = 100 / 50

z = 2

Using the standard z table, at z = 2, the value of P is: (P = proportion)

P (z = 2) = 0.9772

Since we want to find the proportion between 500 and 600, therefore we subtract the two:

P (500 ≥ x ≥ 600) = 0.9772 – 0.5

P (500 ≥ x ≥ 600) = 0.4772

Answer:

Around 47.72% of students have score from 500 to 600.

Step-by-step explanation:

A spinner is divided into 10 equal sections numbered 1 through 10. If the arrow is spun once, what is the probability it will land on a number less than 3?

Answers

well since your options are 3, 2, and 1 and you spin it once, the chances would be 3/10

The probability with the condition of the spinner landing on a number less than 3 is 0.2

What is a conditional probability?

A conditional probability is a probability of an event occuring with a condition that another event had previously occurred. The event in the question is spinning the spinner once while the condition is that the number landed on is less than 3.

The spinner has 10 equal sections numbered 1 through 10.

The conditional probability of landing on a number less than 3 is the same as the probability of landing on either 1 or 2.

There are two sections out of ten that corresponds to numbers less than 3. The probability of landing on a number less than 3 is therefore;

P(Landing on a number less than 3) = P(Landing on 1) + P(Landing on 2)

P((Landing on 1) = 1/10

P(Landing on 2) = 1/10

P(Landing on 1) + P(Landing on 2) = (1/10) + (1/10) = 2/10

(1/10) + (1/10) = 2/10 = 0.2

The probability of landing on a number less than 3 is 0.2

Learn more on conditional probability here: https://brainly.com/question/10431517

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The variable z is directly proportional to x and inversely proportional to y. When x is 12 and y is 18 z has the value 2 what is the value of z when x = 19 and y = 22

Answers

[tex]\bf \qquad \qquad \textit{double proportional variation}\\\\ \begin{array}{llll} \textit{\underline{y} varies directly with \underline{x}}\\ \textit{and inversely with \underline{z}} \end{array}\implies y=\cfrac{kx}{z}\impliedby \begin{array}{llll} k=constant\ of\\ variation \end{array}\\\\ -------------------------------\\\\[/tex]

[tex]\bf z=\cfrac{kx}{y}\impliedby \begin{array}{llll} \textit{directly proportional to "x"}\\ \textit{and inversely proportional to "y"} \end{array} \\\\\\ \textit{we also know that } \begin{cases} x=12\\ y=18\\ z=2 \end{cases}\implies 2=\cfrac{k12}{18}\implies \cfrac{2\cdot 18}{12}=k \\\\\\ \boxed{3=k}\qquad thus\qquad \boxed{z=\cfrac{3x}{y}}\\\\ -------------------------------\\\\ \textit{what's "z" when } \begin{cases} x=19\\ y=22 \end{cases}\implies z=\cfrac{3\cdot 19}{22}[/tex]
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