if you’re good with set theory and word problems in math 30 please help with questions 42 and 43 !! real answers only !!

Answer:

59/34

Step-by-step explanation:

5x+2y=7

-2x+6y=9

Multiply the top equation by 3:

15x+6y=21

Subtract the second equation from the first:

17x=12

x=12/17

Plug this back into one of the other equations to find y:

5(12/17)+2y=7

60/17+2y=7

2y=59/17

y=59/34

Hope this helps!

Answer:

£472.92

Step-by-step explanation:

£2100(0.78)^6

Answer:

This can be done in 120 ways.

Step-by-step explanation:

Number of arrangments:

The number of arrangments of n elements is given by the following formula:

[tex]A_{n} = n![/tex]

In this question:

7 elements.

Two conditions: Counselor in the third sear, and the camper who has the tendency to engage in food fights in the fourth.

For the other 5 seats, the other 5 campers can be arranged in:

[tex]A_{5} = 5! = 120[/tex]

Ways.

This can be done in 120 ways.

Answer:

I think the answer is -15

Answer:

a. [tex]\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}[/tex]

b. [tex]\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}[/tex]

Step-by-step explanation:

The initial value problem is given as:

[tex]y' +y = 7+\delta (t-3) \\ \\ y(0)=0[/tex]

Applying  laplace transformation on the expression [tex]y' +y = 7+\delta (t-3)[/tex]

to get  [tex]L[{y+y'} ]= L[{7 + \delta (t-3)}][/tex]

[tex]l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}[/tex]

Taking inverse of Laplace transformation

[tex]y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}][/tex]

[tex]L^{-1}[\dfrac{e^{-3s}}{s+1}][/tex]

[tex]L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;[/tex]

[tex]L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t <3}} \ \ \ \right.[/tex]

[tex]L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t <3}} \ \ \ \right.[/tex]

[tex]= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t<3}} \right.[/tex]

= [tex]e^{-(t-3)} u (t-3)[/tex]

Recall that:

[tex]y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}][/tex]

Then

[tex]y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)[/tex]

[tex]y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)[/tex]

[tex]\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}[/tex]

Corrected Question

Given a triangle with two sides 4.1 cm and 6.7cm and an Included angle of [tex]35^\circ[/tex]. Use the Law of Cosines to determine the length of the third side of the triangle.

Answer:

(D) 4.1cm

Step-by-step explanation:

Given a triangle with two sides 4.1 cm and 6.7 cm and an included angle of [tex]35^\circ[/tex].

Using Law of Cosines

[tex]a^2=b^2+c^2-2bc\cos A\\a^2=4.1^2+6.7^2-2*4.1*6.7\cos 35^\circ\\a^2=16.6958\\a=\sqrt{16.6958}\\a=4.09\\a \approx 4.1$ cm[/tex]

The correct option is D.

Answer:

The correct answer to the following question will be "5040".

Step-by-step explanation:

Given:

The number of directors,

n = 10

and they select on 4 peoples, then

Number of ways will be:

⇒  [tex]10_{P}_{4}[/tex]

⇒  [tex]\frac{10!}{10-4!}[/tex]

⇒  [tex]\frac{10!}{6!}[/tex]

⇒  [tex]\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1!}{6\times 5\times 4\times 3\times 2\times 1!}[/tex]

⇒  [tex]5040[/tex]

Answer:

96

Step-by-step explanation:

Exterior angles add up to 360

360 - 134-130 = 96

x = 96

Answer:

Step-by-step explanation:

The general form representing limit of a function is expressed as shown below;

[tex]\lim_{h \to a} f(h)[/tex] where a is the value that h will take and use in the function f(h). It can be expressed in words as limit of function f as h tends to a. Comparing the genaral form of the limit to the limit given in question [tex]\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}[/tex], it can be seen that a = 0 and f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]

Taking the limit of the function

[tex]\lim_{h \to 0} \frac{\sqrt{9+h} -3}{h}\\= \frac{\sqrt{9+0}-3 }{0}\\= \frac{0}{0}(indeterminate)[/tex]

Applying l'hopital rule

[tex]\lim_{h \to 0} \frac{\frac{d}{dh} (\sqrt{9+h} - 3)} {\frac{d}{dh} (h)}\\= \lim_{h \to 0} \frac{1}{2} (9+h)^{-1/2} /1\\=\frac{1}{2} (9+0)^{-1/2}\\= \frac{1}{2} * \frac{1}{\sqrt{9} } \\= 1/2 * 1/3\\= 1/6[/tex]

Answer:

Step-by-step explanation:

  The recommended number of volunteers is five (5)

  Since the the distance of the race is 5km,

and the safety committees recommends 1 volunteer per kilometre.

Hence ten (10) volunteers is more than enough

Answer:

you could buy a pair of shoes and a belt still have 95 dollars to spend

Answer:

y-10=5(x-15)

Step-by-step explanation:

Slope =5

15-10/16-15=5

Then plug in 10 as the y1 and 15 as y2.

Answer:

is 3

Step-by-step explanation:

because to find the area for a ractangle you have to multiply LxW and 12x3=36

Answer:

[tex]=\left(x+4\right)\left(x-9\right)[/tex]

Step-by-step explanation:

[tex]x^2-5x-36\\\mathrm{Break\:the\:expression\:into\:groups}\\=\left(x^2+4x\right)+\left(-9x-36\right)\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+4x\mathrm{:\quad }x\left(x+4\right)\\\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9x-36\mathrm{:\quad }-9\left(x+4\right)\\=x\left(x+4\right)-9\left(x+4\right)\\\mathrm{Factor\:out\:common\:term\:}x+4\\=\left(x+4\right)\left(x-9\right)[/tex]

Answer:

Its D

Step-by-step explanation:

becasue the X is not repeating. the other are.

Complete Question:

Ellen is opening. Cookie shop eat a local middle school . She randomly surveys students to determine The types of cookies they would buy at the cookie shop the results of the surveys are below based on the survey. Which statement is not true?

A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

B) 22½% of the students surveyed would buy a sugar cookie.

C) If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.

D) Half of the students prefer chocolate chips or oatmeal raisin cookies.

(Table showing the results of her survey is in the attachment below)

Answer:

A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

Step-by-step Explanation:

STEP 1:

In order to ascertain which of the statements given in the options that is NOT TRUE, let's express the given number of students that would buy the different types of cookies in their PROPORTIONS AND PERCENTAGES in the survey result.

Thus:

Type of Cookie==> no of students => Proportion (no of students of a cookie type ÷ total no of students) => % (each proportion for a cookie type)

Chocolate Chip==> 25 => 0.3125 (25÷80) => 31.25% (0.3125 × 100)

Oatmeal Raisin==> 15 => 0.1875 (15÷850) => 18.75% (0.1875 × 100)

Peanut Butter== 22 => 0.275 (22÷80) => 27.5% (0.275 × 100)

Sugar ==> 18 => 0.225 (18÷80) => 22.5% (0.225 × 100)

STEP 2:

Next step is to consider each statement given in the question to see if they are true or not.

==>OPTION A: If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

To find out if this is true, multiply the proportion of students who would be buy a sugar cookie (0.255) by 160 = 0.255 × 160 = 36.

If 160 buy a cookie, approximately 36 students would buy a sugar.

Option A IS NOT TRUE.

OPTION B: 22½% of the students surveyed would buy a sugar cookie.

From our calculation in STEP 2, we have:

Sugar ==> 18 => 0.225 (18÷80) => 22.5% (0.225 × 100).

Option B is TRUE. 22.5% of the students would go for sugar cookie.

OPTION C: If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.

Number of students that would buy the peanut butter if 240 students were to get a cookie = proportion of students that opt for peanut butter cookie in the survey (0.275) × 240 = 0.275 × 240 = 66.

Option C is TRUE.

OPTION D: D) Half of the students prefer chocolate chips or oatmeal raisin cookies.

Proportion of students who prefer chocolate chips or oatmeal raisin cookies = 0.3125 + 0.1785 = 0.491

This is approximately 0.5 = ½ of the total number of students.

Option D is TRUE.

Therefore, we can conclude that the statement, "A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie" is NOT TRUE.

The number of houses are 180, and the number of flats are 100.

It is given that the number of houses and the number of flats ratio is 9:5 the number of flats and the number of bungalows ratio is 10:3.

It is required to find the number of houses in the village if the number of bungalows is 30.

Fraction number consists of two parts one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

The ratio of the number of houses and the number of flats:

= 9:5   and

The ratio of the number of flats and the number of bungalows :

=10:3

It means we can write the ratio of the number of houses and the number of flats = 18:10

And the ratio of the number of:

Houses : Flats : Bungalows = 18:10:3

But the number of bungalows are 30.

Then the ratios are:

180:100:30

Thus, the number of houses are 180, and the number of flats are 100.

Learn more about the fraction here:

brainly.com/question/1301963

Answer:

In a relation h(x) = y, the values of x are the domain and the values of y are the range.

Here the domain is:

D = {-6, -1, 7}

The range is:

R = {-9, 0 , 9}

Now, a relation f(x) = y is a function only if for every x in the domain we have only and only one value in the range.

Here we can see that for the value -1 in the domain we have two different values in the range:

(-1, 0) and (-1, 9)

So this can not be a function.

(If you want to take the variable as y, you also have that the value y = 0 leads to two different values in x, so this cant be a function either)

Answer:

The correct answer is I am 95% sure that this interval includes the population mean arm length.

Note: Kindly find an attached image or copy of the complete question to this solution below:

Sources: The complete question was researched from Quizlet and Course hero sites.

Step-by-step explanation:

Solution

From the given question, i will say that I am 95 % confident interval that this interval include the population mean arm length is correct because using sample mean and confidence level we can obtain an interval which gives us the certain level of confidence that population mean is within this interval and here we are 95% confident that population mean is within this interval.

Answer:

1.5 hr and 3.5 hr

Step-by-step explanation:

Time in the first part= x

Time in the second part= 5 - x

Answer:

undefined

Step-by-step explanation:

y = -8 is a horizontal line with a slope of zero

A line that is perpendicular is a vertical line, which has a slope of undefined

Answer:

A) sample mean = $1.36 million

B) standard deviation = $0.9189 million

C) variance confidence interval = ($0.40 million, $2.81 million)

D) standard deviation confidence interval = ($1.93 million , $0.79 million)

*since the sample size is very small, the confidence interval is not valid.

Step-by-step explanation:

samples:

sample mean = $1.36 million

the standard deviation:

variance 0.8444

standard deviation = √0.8444 = 0.9189

in order to calculate the confidence interval for the population variance we are going to use a chi-square distribution with 2.5% on each tail ⇒ table values 2.7004 and 19.023 enclose 95% of the distribution.

[(n - 1) x variance] / 2.7004 = (9 x 0.8444) / 2.7004 = 2.81

[(n - 1) x variance] / 19.023 = (9 x 0.8444) / 19.023 = 0.40

95% confidence interval = mean +/- 1.96 standard deviations/√n:

$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million

$1.36 million - $0.57 million = $0.79 million

Answer:

Answer D is correct

Answer:

[tex]V=122.5in^3[/tex]

Step-by-step explanation:

The volume of a right circular cone is given by:

[tex]V=\frac{\pi r^2h}{3}[/tex]

where r is the radius of the circle and h is the height of the cone, and [tex]\pi[/tex] is a constant [tex]\pi=3.1416[/tex].

According to the problem the height is:

[tex]h=8.1 in[/tex]

and we don't have the radius but we have the diameter, which is useful to find it. We just divide the diameter by 2 to find the radius:

[tex]r=\frac{d}{2}=\frac{7.6in}{3}=3.8in[/tex]

Now, we can find the volume by substituting all the known values:

[tex]V=\frac{\pi r^2h}{3}[/tex]

[tex]V=\frac{(3.1416)(3.8in)^2(8.1in)}{3} \\\\V=\frac{(3.1416)(14.44in^2)(8.1in)}{3} \\\\V=\frac{367.454in^3}{3} \\\\V=122.485[/tex]

Rounding the volume to the nearest tenth of cubic inch we get:

[tex]V=122.5in^3[/tex]

Answer:

Step-by-step explanation:

For focus-to-vertex distance "p", the equation of a parabola with vertex (h, k) can be written as ...

  y = 1/(4p)(x -h)^2 +k

Comparing this to your equation, we see that ...

  1/(4p) = 1

  h = -2

  k = -3

Solving for p, we find ...

  1/(4p) = 1

  1/4 = p . . . . . multiply by p

The parabola opens upward, so this means the focus is 1/4 unit above the vertex, and the directrix is 1/4 unit below the vertex.

Answer:

The focus is at (–2,–212) and the directrix is at y = –312.

Step-by-step explanation:

Find the focus and directrix of the parabola y=12(x+2)2−3.

got the answer right in the test.

Linear equations are typically organized in slope-intercept form:

[tex]y=mx+b[/tex]

Perpendicular lines have slopes that are negative reciprocals.

We're given:

1) Determine the slope

Let's first rearrange this equation into slope-intercept form:

[tex]4x-5y=-10\\-5y=-4x-10\\\\y=\dfrac{4}{5}x+2[/tex]

Notice how [tex]\dfrac{4}{5}[/tex] is in the place of m in y = mx + b. This is the slope of the give line.

Since perpendicular lines are negative reciprocals, we know the slope of the other line is [tex]-\dfrac{5}{4}[/tex]. Plug this into y = mx + b:

[tex]y=-\dfrac{5}{4}x+b[/tex]

2) Determine the y-intercept

We're also given that the line passes through (6,3). Plug this point into our equation and solve for b:

[tex]y=-\dfrac{5}{4}x+b\\\\3=-\dfrac{5}{4}(6)+b\\\\b=3+\dfrac{5}{4}(6)\\\\b=\dfrac{21}{2}[/tex]

Plug this back into our original equation:

[tex]y=-\dfrac{5}{4}x+\dfrac{21}{2}[/tex]

[tex]y=-\dfrac{5}{4}x+\dfrac{21}{2}[/tex]

Answer:

Step-by-step explanation:

Let X denote the dimension of the part after grinding

X has normal distribution with standard deviation [tex]\sigma=0.002 in[/tex]

Let the mean of X be denoted by [tex]\mu[/tex]

there is an upper specification of 3.150 in. on a dimension of a certain part after grinding.

We desire to have no more than 3% of the parts fail to meet specifications.

We have to find the maximum [tex]\mu[/tex] such that can be used if this 3% requirement is to be meet

[tex]\Rightarrow P(\frac{X- \mu}{\sigma} <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{\sigma} )\leq 0.03\\\\ \Rightarrow P(Z <\frac{3.15- \mu}{0.002} )\leq 0.03[/tex]

We know from the Standard normal tables that

[tex]P(Z\leq -1.87)=0.0307\\\\P(Z\leq -1.88)=0.0300\\\\P(Z\leq -1.89)=0.0293[/tex]

So, the value of Z consistent with the required condition is approximately -1.88

Thus we have

[tex]\frac{3.15- \mu}{0.002} =-1.88\\\\\Rrightarrow \mu =1.88\times0.002+3.15\\\\=3.15[/tex]

Answer:

the volume of the box is increasing

dV = +310,000 ft^3/s

Step-by-step explanation:

Volume of a rectangular box with side a,b and c can be expressed as;

V = abc

The change in volume dV can be expressed as;

dV = d(abc)/da + d(abc)/db + d(abc)/dc

dV = bc.da + ac.db + ab.dc ......1

Given:

a= 150 feet,

b= 80 feet, and

c= 50 feet

ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec

da = +100 feet/s

db = -20 feet/s

dc = +5 feet/s

Substituting the values into the equation 1;

dV = (80×50×+100) + (150×50×-20) + (150×80×+5)

dV = +400000 - 150000 + 60000 ft^3/s

dV = +310,000 ft^3/s

Since dV is positive, the volume of the box is increasing at that instant.

Answer:

Evaluate f(7)

[tex] f(7) = 240(0.7)^7 = 19.765[/tex]

Determine x when f(×)=120

[tex] 120 = 240 (0.7)^x[/tex]

We can derive both sides by 240 and we got:

[tex] 0.5 = 0.7^x[/tex]

Now we can apply natural log on both sides and we got:

[tex] ln(0.5)= x ln(0.7)[/tex]

And if we solve for the value of x we got:

[tex] x =\frac{ln(0.5)}{ln(0.7)}= 1.943[/tex]

And then the value of x = 1.943

Step-by-step explanation:

We have the following function given:

[tex]f(x) = 240(0.7)^x [/tex]

Evaluate f(7)

And we want to find [tex] f(7) [/tex] so we just need to replace x=7 and we got:

[tex] f(7) = 240(0.7)^7 = 19.765[/tex]

Determine x when f(×)=120

And for the second part we want to find a value of x who satisfy that the function would be equal to 120 and we can set up this:

[tex] 120 = 240 (0.7)^x[/tex]

We can derive both sides by 240 and we got:

[tex] 0.5 = 0.7^x[/tex]

Now we can apply natural log on both sides and we got:

[tex] ln(0.5)= x ln(0.7)[/tex]

And if we solve for the value of x we got:

[tex] x =\frac{ln(0.5)}{ln(0.7)}= 1.943[/tex]

And then the value of x = 1.943