ASAPP help me with this question

The angles and length of the triangle area as follows:

The sum of angles in a triangle is 180 degree.

Therefore,

∠DBC = 180 - 29 - 90 = 61°

Hence,

∠A = 180 - 29 - 61 - 55 = 35°

∠CBD = 61°

Using trigonometric ratios,

sin 55 = opposite / hypotenuse

sin 55 = AD / 40

AD = 40 sin 55

AD = 32.7660817716

AD = 32.77 units

cos 55 = adjacent / hypotenuse

cos 55 = BD / 40

BD = 40 cos 55

BD = 22.943057454

BD = 22.94 units

sin 29 = 22.94 / BC

BC = 22.94 sin 29

BC = 11.1215326885

BC = 11.12 units

cos 29 = CD / 11.12

CD = 11.12 cos 29

CD = 9.72577114339

CD = 9.73 units

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Answer: 8.7 the mean is 20

Step-by-step explanation:

Tthe inequality that describes this graph is y ≤ 1/3x - 4/3

The graph that completes the question is added as an attachment

From the attached graph, we have the following points

(0, -1.3) and (3, -0.3)

The slope is calculated as:

m = (y2 - y1)/(x2 - x1)

Substitute the known values in the above equation

m = (-0.3 + 1.3)/(3 - 0)

Evaluate

m = 1/3

The equation is then calculated as:

y = m(x - x1) + y1

This gives

y = 1/3(x - 0) - 1.3

Evaluate

y = 1/3x - 4/3

From the graph, we have the following highlights:

The first highlight above implies, the inequality can be any of ≥ and ≤

While the second highlight above implies, the inequality is ≤

Hence, the inequality that describes this graph is y ≤ 1/3x - 4/3

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Answer:

250 - 6 - 3n

Step-by-step explanation:

Let n equal the number of students.

Answer:

[tex]y=\frac{-3}{4}x-3[/tex]

Step-by-step explanation:

This is a linear graph so the equation will be in the format of y=mx+b.

b, the y-intercept, will be -3, as that is where the line crosses the y-axis.

m, the slope, is [tex]\frac{-3}{4}[/tex], calculated from the change in y over the change in x from points (-4,0) to (0,-3).

Now substitute the values in our equation:

[tex]y=\frac{-3}{4}x-3[/tex]

Answer:

y= -3/4+[-3]

Step-by-step explanation:

well, keeping in mind that a year has 365 years, so let's see

[tex]\stackrel{Oct}{10}~~ + ~~\stackrel{Nov}{30}~~ + ~~\stackrel{Dec}{31}~~ + ~~\stackrel{Jan}{31}~~ + ~~\stackrel{Feb}{21}\implies 123~days[/tex]

so the 3600 were borrowed for only 123 days of a year, that'd be 123/365 of a year, thus

[tex]~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$3694.63\\ P=\textit{original amount deposited}\dotfill & \$3600\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &\frac{123}{365} \end{cases}[/tex]

[tex]3694.63=3600[1+(\frac{r}{100})(\frac{123}{365})]\implies \cfrac{3694.63}{3600}=1+\cfrac{123r}{36500} \\\\\\ \cfrac{3694.63}{3600}-1=\cfrac{123r}{36500}\implies \left( \cfrac{36500}{123} \right)\left( \cfrac{3694.63}{3600}-1 \right)=r\implies \stackrel{\%}{7.8}\approx r[/tex]

7.8% is the simple interest rate was charged if Robin repaid $3,694.63.

Simple interest is based on the principal amount of a loan or the first deposit in a savings account.

A=P(1+rt)

A is final amount, P is principle amount, r is rate of interest and t is time.

We know that in a year we have 365 days.

But the time here is from oct 21 to feb 21, which has 123 days.

so the 3600 were borrowed for only 123 days of a year, that'd be 123/365 of a year, thus

Apply these values in formula

A=P(1+rt)

3694.63=3600(1+r/100(123/365))

3694.63=3600(1+123r/36500)

3694.63/3600=1+123r/36500

3694.63/3600-1=123r/36500

(3694.63-3600)/3600=123r/36500

94.63/3600=123r/36500

Apply cross multiplication

3453995=442800r

7.80=r

Hence 7.8% is the simple interest rate was charged if Robin repaid $3,694.63.

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Option A. The total share expense that the company would share would be given as 512,000

These are the necessary expenses that are needed for the smooth functioning of a particular business that are not within the confinement of the O and M agreement. It has to do with shared facilities.

The total number of the employees that are knwon to satistfy condition are given as

800 * 0.8

= 640

The options that are estmated that would be exercised

This is given as the employees * share option

= 640×50

=32000.

The total shae for the company would be gotten as

= 32000× $16

This gives us $512,000.

Hence it can be concluded that the total share of the company if they have 80 percent meeting the condition is $512000.

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Answer:

your question is not correct i can not find the answer

Usando la distribución hipergeométrica,  la probabilidad de que los alumnos del colegio A obtengan los dos premios (no hay empates), es: 0.2 = 20%.

La fórmula es:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Los parámetros son:

Los valores de los parámetros son:

N = 6, k = 3, n = 2.

La probabilidad de que los alumnos del colegio A obtengan los dos premios (no hay empates), es P(X = 2), por eso:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,6,2,3) = \frac{C_{3,2}C_{3,0}}{C_{6,2}} = 0.2[/tex]

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Answer: 8 x 10^2

Step-by-step explanation:

scientific notation has to be greater than 1, less than 10

therefore:

0.08 = 8 x 10^2

The best conjecture for Charlie to develop is that an altitude drawn to the base of the isosceles triangle creates the smaller congruent triangles by the HL postulate.

It should be noted that conjecture simply means a conclusion that's proffered on a tentative basis without any proof.

From the information, Charlie used construction paper, scissors and a ruler to construct an isosceles triangle with a pair of sides each 3 inches long and an acute angle on top.

Here, the best conjecture for Charlie to develop is that an altitude drawn to the base of the isosceles triangle creates the smaller congruent triangles by the HL postulate.

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y=x²+(3+1)x-3

y=x²+3x+1x-3

y=x(x+3)-1(x+3)

y=(x-1)(x+3)

When we divide [tex]x^{3} +6x^{2} +3x+1[/tex] by (x-2) we will get the quotient be [tex]x^{2} +8x+19[/tex] and remainder be 39.

Given two expressions be [tex]x^{3} +6x^{2} +3x+1[/tex] and (x-2).

We are required to divide the first expression by second expression.

Division means distributing parts of something. The number which is being divided is known as quotient.Divisor is a number which divides the number.

Expressions refers to the combination of numbers, fractions, coefficients, determinants, indeterminants. It expresses some relationship or show equation of line.

We know that  relationship between quotient, divisor, divident and remainder is as under:

Dividend=Divisor*Quotient+Remainder

[tex]x^{3} +6x^{2} +3x+1[/tex]=(x-2)*([tex]x^{2} +8x+19[/tex])+39

Quotient=[tex]x^{2} +8x+19[/tex]

Remainder=39

Hence when we divide [tex]x^{3} +6x^{2} +3x+1[/tex] by (x-2) we will get the quotient be [tex]x^{2} +8x+19[/tex] and remainder be 39.

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At the growth rate of [tex]6.7\%[/tex], the population of the town after 4 years can be represented by the equation, [tex]A=1.23\times 10\times(1.067)^4[/tex].

Here, the initial population is [tex]P=1.23\times 10[/tex] and the growth rate is [tex]r=6.7\%[/tex].

So the population after [tex]t=4[/tex] years will be:

[tex]A=P(1+\frac{r}{100})^t\\\Longrightarrow A=1.23\times 10\times(1+\frac{6.7}{100})^4\\\Longrightarrow A=1.23\times 10\times(1.067)^4[/tex]

Therefore, at the growth rate of [tex]6.7\%[/tex], the population of the town after 4 years can be represented by the equation, [tex]A=1.23\times 10\times(1.067)^4[/tex].

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Answer:

A)

Step-by-step explanation:

3+3=6 6>5

3+5=8 3<8

A)

Answer:

Step-by-step explanation:

2x + 5y = 8

      -5    -5

2x = 8 - 5y

divide both sides by 2

[tex]\frac{2x}{2} = \frac{8-5y}{2}[/tex]

divide by 2 undoes the multiplication by 2

x = [tex]\frac{8- 5y}{2}[/tex]

divide 8 - 5y by 2

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

Rewriting the left-hand side as follows,

[tex](2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40[/tex]

The equation 15x + 5y = 0  is a direct variation and the constant of variation is -3 while the equation -5x + 8y = 3 is neither a direct variation nor an inverse variation

As a general rule, we have the following forms of equation:

Direct variation: y =kx

Inverse variation: y = k/x

Where k represents the constant of variation

The equations 15x + 5y = 0 can be rewritten as:

5y = -15x

Divide by 5

y = -3x

The above equation take the form y = kx

This means that k = -3

Hence, the equation 15x + 5y = 0  is a direct variation and the constant of variation is -3

However, the equation -5x + 8y = 3 do not take any of the given forms

Hence, the equation -5x + 8y = 3 is neither a direct variation nor an inverse variation

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Yes, companies that provide mobile wifi hotspots can see your history.

A Wi-Fi hotspots can be defined as  the hotspots that makes it possible for other mobile user connected to a wifi hotspots to browse or to connect to the internet.

For a mobile user to connect to the internet through the mobile wifi hotspot, the mobile hotspot wifi owner must have giving access to mobile user to browse .

Hence, companies that provide mobile wifi hotspots can see your history of the site you visit.

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The number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

Note that the formula for combination is given as;

Combination = [tex]\frac{n!}{r!(n-r)!}[/tex]

From the information we have that;

There are 28 questions in the pool

The test should have a total of 10 questions;

6 difficult , 10 average and 12 easy questions

We are asked to determine the combination of;

3 difficult questions

4 average questions

3 easy questions

6C3 = [tex]\frac{6!}{3!(6-3)!}[/tex]

6C3 = [tex]\frac{720}{36}[/tex]

6C3 = 20

10C4 = [tex]\frac{10!}{4!(10-4)!}[/tex]

10C4 = [tex]\frac{3628800}{17280}[/tex]

10C4 = 210

12C4 = [tex]\frac{12!}{4!(12-4)!}[/tex]

12C4 = 220

The number of ways of constructing the questions is

= 20 × 210 × 220

= 924, 000 ways

Thus, the number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

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Answer:

Step-by-step explanation:

Converting the portions to decimal, the grade that has the greatest portion of students is: 3rd grade.

The bigger the digit that is close to the decimal point, the bigger the number given.

Also, percentage can be converted to decimal by dividing the given percent by 100. Percent means over 100.

Fractions can also be converted to decimals by simply dividing the numerator by the denominator.

To compare the portion of students interested in playing an instrument in each grade, let all the portions be in decimals.

Converting 25/36 to decimal, we have: 0.694 (3rd grade)

Converting 61.24% to decimal, we have: 61.24/100 = 0.6124 (4th grade)

From the rest of the portions given, 0.694 is greater than the others.

Therefore, the grade that has the greatest portion of students is: 3rd grade.

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Answer:

12 and 14

Step-by-step explanation:

12 * 14 = 168

Answer:

12 & 14

Step-by-step explanation:

the work: x = first even number, (x+2) = second even number

The equation

[tex]x(x+2)=168[/tex]

[tex]x^{2} +2x=168[/tex]

[tex]x^{2} +2x-168=0[/tex]

Factoring

[tex](x-12)(x+14)=0[/tex]

first factor

[tex]x-12=0\\x=12[/tex]

Second factor

[tex]x+14=0\\x=-14[/tex]

The solution is the positive number

x = 12     is the first even number

x + 2 = 12 + 2 = 14    is the second even number

Hope this helps

By the chain rule,

[tex]g(x) = f(4 - x^2) \\\\ \implies g'(x) = -2x f'(4 - x^2) \\\\ \implies g''(x) = -2 f'(4 - x^2) + 4x^2 f''(4 - x^2)[/tex]

Observe that

[tex]4 - x^2 = 1 \implies x^2 = 3 \implies x = \pm\sqrt3[/tex]

Then

[tex]g''(\sqrt3) = -2 f'(1) + 4\left(\sqrt3\right)^2 f''(1) = -2(-5) + 4(3)(-1) = \boxed{-2}[/tex]

Using translation concepts, the transformations are described as follows:

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

In this problem, the parent function is:

[tex]f(x) = 2^x[/tex]

The transformed function is:

[tex]g(x) = -2^x + 4[/tex]

We have that the transformations in this problem can be described as follows:

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Answer:

3 hours

Step-by-step explanation:

When the ships meet each other, that means the the distance travelled for both ships are the same, that is where the 2nd ship will overtake the ship.

Distance = Speed * Time

So, given information from the question, and letting x be the time in hours (from when the 1st ship departs), we can deduce:

Distance travelled of 1st ship = Distance travelled of 2nd ship

30x = 40(x-1)

30x = 40x - 40

-10x = - 40

x = [tex]\frac{-40}{-10}[/tex]

= 4

Since we are calculating based on the 2nd ship,

Time taken for 2nd ship to overtake first ship = x - 1 = 4 - 1 = 3 hours.

The decision is to fail to reject the null given that p value is not ≤ 0.05.

The hypothesis

H0 = p = 0.63

H1 = p < 0.63

α = 0.05

sample proportion = 71/125 = 0.568

x = 71

n = 125

standard error of the proportion

√0.63(1-0.63)/125

= 0.0431

The null hypothesis follows a standard normal distribution. This is a left tailed test.

We are to reject the null if the p value is less than 0.05

p < 0.05

This probability test is a z probability test.

Z test = 0.568 - 0.63 / 0.0431

test statistic = -1.436

-Z0.05 =

Critical value =  -1.645

p(z < -1.436)

p value = 0.0755

The decision would be to fail to reject the null hypothesis. The reason would be due to the fact that p value is greater than significance.

P-value is not ≤ α 0.05

0.0755 > 0.05

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Solve the second equation for [tex]t[/tex], then substitute it into the first equation.

[tex]y = t + 2 \implies t = y-2[/tex]

[tex]x = t^2 - 3 \implies \boxed{x = (y-2)^2 - 3}[/tex]

Answer:

sorry i thought i knew it

Step-by-step explanation:

Answer:

3rd option

Step-by-step explanation:

using the rules of exponents

[tex]\frac{a^{m} }{a^{n} }[/tex] = [tex]a^{(m-n)}[/tex] : nm > n

[tex]\frac{a^{m} }{a^{n} }[/tex] = [tex]\frac{1}{a^{(n-m)} }[/tex] : n > m

[tex]\frac{6a^2b^{-2} }{8a^{-3b^3} }[/tex] ← separate the variables

= [tex]\frac{6}{8}[/tex] × [tex]\frac{a^2}{a^{-3} }[/tex] × [tex]\frac{b^{-2} }{b^3}[/tex]

= [tex]\frac{3}{4}[/tex] × [tex]a^{2-(-3)}[/tex] × [tex]\frac{1}{b^{3-(-2)} }[/tex]

= [tex]\frac{3}{4}[/tex] × [tex]a^{2+3}[/tex] × [tex]\frac{1}{b^{3+2} }[/tex]

= [tex]\frac{3}{4}[/tex] × [tex]a^{5}[/tex] × [tex]\frac{1}{b^{5} }[/tex]

= [tex]\frac{3a^{5} }{4b^{5} }[/tex]