Compute the area of triangle, if x equals 4 more than 6. A) 10 B) 50 C) 100 D) 200 E) 400

Answer:

24.06% of total sales

Step-by-step explanation:

Total sales, which were $170,000 originally, are expected to grow by 10%. The expected value of total sales this year is:

[tex]S=\$170,000*1.10 = \$187,000[/tex]

If Basil sales remain at the same value of $45,000, the percentage of sales corresponding to basil is:

[tex]B=\frac{\$45,000}{\$187,000}*100\%\\B=24.06\%[/tex]

Therefore, basil will correspond to 24.06% of total sales.

Basil sales will represent approximately 24.06% of Scarborough Faire Herb Farm's projected total sales this year, given that total sales are forecasted to increase by 10% and basil sales remain unchanged.

The student's question concerns the percentage of total sales that basil sales will represent for Scarborough Faire Herb Farm in the current year, assuming a 10% increase in total sales from the previous year and unchanged basil sales. Firstly, we calculate the projected total sales for this year by increasing last year's total sales by 10%. The calculation is as follows:

Since the basil sales are to remain the same as last year ($45,000), we now calculate what percentage this represents of the projected total sales for this year:

Thus, basil sales will account for approximately 24.06% of the Scarborough Faire Herb Farm's total sales this year.

Answer:

The cost of goods sold is 65,850

Step-by-step explanation:

FIFO Perpetual chart is attached.

FIFO Perpetual chart shows purchases, sales and balance of  the period.

The cost of goods sold is:

3,090 units x $5=$15,450

6,300 units x $8=$50,400

Total=65,850

Answer:

The average rate of change as follows of I is -0.185.

Step-by-step explanation:

The relation between voltage V, current I, and resistance R in a circuit, according to the Ohm's law is:

[tex]V = IR[/tex]

It is provided that:

V = 20 V

The interval between which R varies is, R = 8 to R = 8.1.

Compute the value of I as follows:

[tex]V=IR\Rightarrow I=\frac{I}{R}\Rightarrow I=\frac{20}{R}[/tex]

Compute the average rate of change as follows of I as follows:

[tex]\frac{\delta I}{\delta R}=\frac{(I\times8.1)-(I\times8)}{8.1-8}[/tex]

    [tex]=\frac{1}{0.1}[\frac{12}{8.1}-\frac{12}{8}][/tex]

    [tex]=\frac{12}{0.1}[\frac{8-8.1}{64.8}][/tex]

    [tex]=-0.1852[/tex]

Thus, the average rate of change as follows of I is -0.185.

Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

[tex]\frac{dB}{dt} = rB-k[/tex]

[tex]\Rightarrow \frac{dB}{dt} - rB=-k[/tex].......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor [tex]=e^{\int p(t) dt[/tex]

                                     [tex]=e^{\int (-r)dt[/tex]

                                     [tex]=e^{-rt}[/tex]

Multiplying the integrating factor the both sides of equation (1)

[tex]e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}[/tex]

[tex]\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt[/tex]

Integrating both sides

[tex]\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt[/tex]

[tex]\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C[/tex]        [ where C arbitrary constant]

[tex]\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}[/tex]

Initial condition B=7864 when t =0

[tex]\therefore 7864= \frac{k}{r} - Ce^0[/tex]

[tex]\Rightarrow C= \frac{k}{r} -7864[/tex]

Then the general solution is

[tex]B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}[/tex]

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

[tex]B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}[/tex]

[tex]\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0[/tex]

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

                                                 =$1573.17

The payment rate that is required to pay off the loan in 3 years is approximately 2949.51 dollars/year. The interest paid during the 3-year period is about 984.53 dollars.

To solve this problem, the formula for continuously compounded interest should be used which is A = P * e^(rt), where A is the value of the investment at a future time t, P is the principal amount, r is the annual interest rate, and t is the time the money is invested or borrowed for.

First, set up the equation: 7864 = k * (1/(0.13)) * (e^(0.13 * 3) - 1), then solve for k: k = 7864 * (0.13) / (e^(0.13 * 3) - 1) ≈ 2949.51 dollars/year.

Using the formula A = P * e^(rt), the total amount paid is A = 2949.51 * 3 = 8848.53 dollars. The interest paid during the 3-year period is calculated by subtracting the loan amount from the total amount paid, that is 8848.53 - 7864 = 984.53 dollars.

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The point between (4,16) and (16,16) is (10,16) as calculated using the midpoint formula. This point is exactly halfway between the given points.

To determine a point between the two points (4,16) and (16,16), we need to calculate the midpoint.

[tex]M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)[/tex]

[tex]M = \left(\frac{4 + 16}{2}, \frac{16 + 16}{2}\right) = (10, 16)[/tex]

Therefore, the point that lies between (4,16) and (16,16) is (10,16).

Answer:

w = $1 or $10 or $25

Probability distribution of w = ($1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $10,$1  $10,$1  $10,$10  $10,$25  $10,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25)

Step-by-step explanation:

Since the winner gets the larger amount of the two slips picked, random variable 'w' which is the amount awarded could be $1 or $10 or $25.

The probability distribution of 'w' is the values that the statistic takes on. Which could be: $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $10,$1  $10,$1  $10,$10  $10,$25  $10,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25

For [tex]0\le t\le1[/tex], we have

[tex]y'+6y=1\implies e^{6t}y'+6e^{6t}=(e^{6t}y)'=e^{6t}\implies y=\dfrac16+Ce^{-6t}[/tex]

Given that [tex]y(0)=0[/tex], we have

[tex]0=\dfrac16+C\implies C=-\dfrac16[/tex]

so that

[tex]y=\dfrac16(1-e^{-6t})[/tex]

For [tex]t>1[/tex] (I think you mistakenly wrote [tex]t>0[/tex], which overlaps with the first definition of [tex]g(t)[/tex]), we have

[tex]y'+6y=0\implies e^{6t}y'+6e^{6t}y=(e^{6t}y)'=0\implies y=Ke^{-6t}[/tex]

We want this to be a continuation of the previously found solution [tex]y[/tex] at [tex]t=1[/tex], which means we need to pick [tex]K[/tex] such that

[tex]\dfrac16(1-e^{-6})=Ke^{-6}\implies K=\dfrac16(e^6-1)[/tex]

Then the solution to the IVP is

[tex]y(t)=\begin{cases}\frac16(1-e^{-6t})&\text{for }0\le t\le1\\\frac{e^6-1}6e^{-6t}&\text{for }t>1\end{cases}[/tex]

Alternatively, we can get around treating [tex]g(t)[/tex] piecemeal and resorting to the Laplace transform. Write

[tex]g(t)=\begin{cases}1&\text{for }0\le t\le1\\0&\text{for }t>1\end{cases}=u(t)-u(t-1)[/tex]

where

[tex]u(t-c)=\begin{cases}0&\text{for }t<c\\1&\text{for }t\ge c\end{cases}[/tex]

is the unit step function.

Take the Laplace transform of both sides of the ODE:

[tex]y'+6y=g(t)\overset{\text{L.T.}}{\implies}(sY-y(0))+6Y=\mathcal L_s\{g(t)\}[/tex]

where [tex]Y=Y(s)[/tex] is the Laplace transform of [tex]y(t)[/tex].

We have

[tex]\mathcal L_s\{g(t)\}=\displaystyle\int_0^\infty g(t)e^{-st}\,\mathrm dt=\int_0^1e^{-st}\,\mathrm dt=\dfrac{1-e^{-s}}s[/tex]

so that

[tex](s+6)Y=\frac{1-e^{-s}}s\implies Y=\dfrac{1-e^{-s}}{s(s+6)}=\dfrac{1-e^{-s}}6\left(\dfrac1s-\dfrac1{s+6}\right)[/tex]

Taking the inverse transform yields

[tex]y=\dfrac{1-u(t-1)}6-\dfrac{e^{-6t}}6(e^tu(t-1)-1)[/tex]

[tex]y=\dfrac{1-e^{-6t}}6+\dfrac{e^{6-6t}-1}6u(t-1)[/tex]

which is equivalent to the same solution found earlier; for [tex]0\le t\le1[/tex], [tex]u(t-1)=0[/tex], so that [tex]y=\frac{1-e^{-6t}}6[/tex]; for [tex]t>1[/tex], [tex]u(t-1)=1[/tex], and [tex]y=\frac{1-e^{-6t}}6+\frac{e^{6-6t}-1}6=\frac{(e^6-1)e^{-6t}}6[/tex].

The given differential equation needs to be solved separately for two time ranges because of the piecewise-defined function g(t). Solution for the corresponding equations are founded using the techniques of homogeneous equation solutions and the integrating factor method. These solutions are then matched at the point of continuity.

The given differential equation is a first-order linear differential equation of the form y′+p(t)y=g(t). We need to solve this equation considering two cases due to the piecewise definition of g(t).

Case 1: For 0 ≤ t ≤ 1, g(t) = 1. The corresponding homogeneous equation is y' + 6y = 0, with the solution being y(t) = Ce-6t. We find the particular solution using the integrating factor method, yielding y(t) = t/6 - 1/36 + Ce-6t. Substituting the initial condition y(0) = 0 helps us solve for C, giving the final solution for this range as y(t) = t/6 - 1/36.

Case 2: For t > 1, g(t) = 0. The homogeneous solution is the same as in Case 1, but in this case, no particular solution needs to be added, so the solution is y(t) = Ce-6t. The constant is determined by making the function continuous at t=1. We ultimately get y(t) = (1-e-6(t-1))/36.

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Answer:

a). 0.03593, 0.27425

(bl 0.0703

(c) the lifespan of such bulb is less than 210 days

(d) 0.0298.

Step-by-step explanation:

Given that U = 300, sd= 50

Z = X-U/sd

(a) We find the probability

Pr(X<210) = Pr(Z<(210-300)/50)

= Pr(Z<-1.8)

= 0.03593

Pr(X >330) = Pr(Z >(330-300)/50))

= Pr(Z> 0.6)

= 1- PR(Z<0.6) = 1 - 0.72575

Pr(X>330)= 0.27425

(b) Pr(280< X< 330) = Pr( 280 < Z< 380)

= Pr([280-300/50] < Z < [380-300/50])

= Pr(0.4 < Z < 1.6)

= Pr(Z< 1.6) - Pr(Z < 0.4)

=0.72575 - 0.65542

= 0.07033

= 0.0703

(c) the lifespan of such bulbs is less than 210 days

(d) given n = 6, x = 4 ,

Pr(X>330) = 0.27425 = p

q = 1-p = 0.72575

Pr(X=4) = 6C4 × (0.27425)⁴ ×(0.72575)²

Pr(X=4) = 10 × 0.005657 × 0.52671

Pr(X= 4) = 0.029795

Pr(X= 4) = 0.0298

Answer:

The Expected cosy of the builder is $433.3

Step-by-step explanation:

$400 is the fixed cost due to delay.

Given Y ~ U(1,4).

Calculating the Variable Cost, V

V = $0 if Y≤ 2

V = 50(Y-2) if Y > 2

This can be summarised to

V = 50 max(0,Y)

Cost = 400 + 50 max(0, Y-2)

Expected Value is then calculated by;

Waiting day =2

Additional day = at least 1

Total = 3

E(max,{0, Y - 2}) = integral of Max {0, y - 2} * ⅓ Lower bound = 1; Upper bound = 4, (4,1)

Reducing the integration to lowest term

E(max,{0, Y - 2}) = integral of (y - 2) * ⅓ dy Lower bound = 2; Upper bound = 4 (4,2)

E(max,{0, Y - 2}) = integral of (y) * ⅓ dy Lower bound = 0; Upper bound = 2 (2,0)

Integrating, we have

y²/6 (2,0)

= (2²-0²)/6

= 4/6 = ⅔

Cost = 400 + 50 max(0, Y-2)

Cost = 400 + 50 * ⅔

Cost = 400 + 33.3

Cost = 433.3

Answer:

the expected value of the builder’s cost due to waiting for supplies is $433.3

Step-by-step explanation:

Due to the integration symbol and also for ease of understanding, i have attached the explanation as an attachment.

Answer:

Step-by-step explanation: see attachment

Answer:

x=-0.75

Step-by-step explanation:

Combine Like terms

2/3x= -5/2+2

2/3x= -2.5+2

2/3x= -0.5

Multiply both sides by 3

2x= -1.5

Divide both sides by 2

x= -0.75

Answer:

x = -4/3

Step-by-step explanation:

The LCM for the fractions is 6x.

2/3x = -5/2 + 2

Multiply through by 6x

(6x) 2/3x = (6x) -5/2 + (6x)2

4 = -15x + 12x

4 = -3x

x = -4/3

Enjoy Maths!

Answer:

a. 15.9%

b. 1.2%

Step-by-step explanation:

using the normal distribution we have the following expression

[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{5-M}{SD}) \\[/tex]

Where  the first expression in the right hand side is the z-scores, M is the mean of value 3.77 and SD is the standard deviation of value 1.23.

if we simplify and substitute values, we arrive at

[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{a-M}{SD}) \\P(X\geq 5)=P(Z \geq \frac{5-3.77}{1.23})\\P(X\geq 5)=P(Z \geq 1)\\P(X\geq 5)=1- P(Z < 5)\\P(X\geq 5)=1-0.841\\P(X\geq 5)=0.159[/tex]

in percentage, we arrive at 15.9%

b. for the percentage rated the eyes most a 1

[tex]P(X\leq a)=P(\frac{X-u}{\alpha } \leq \frac{a-M}{SD})\\for a=1\\P(X\leq 1)=P(Z \leq \frac{1-3.77}{1.23})\\P(X\leq 1)=P(Z \leq -2.25})\\P(X\leq 1)=0.012\\[/tex]

In percentage we have 1.2%

Answer:

The probability that all 70 messages are transmitted in less than 12 minutes is 0.117.

Step-by-step explanation:

Let X = the transmission time of each message.

The random variable X follows an Exponential distribution with parameter λ = 5 minutes.

The expected value of X is:

[tex]E(X)=\frac{1}{\lambda}=\frac{1}{5}=0.20[/tex]

The variance of X is:

[tex]V(X)=\frac{1}{\lambda^{2}}=\frac{1}{5^{2}}=0.04[/tex]

Now define a random variable T as:

T = X₁ + X₂ + ... + X₇₀

According to a Central limit theorem if a large sample (n > 30) is selected from a population with mean μ and variance σ² then the sum of random variables X follows a Normal distribution with mean, [tex]\mu_{s} = n\mu[/tex] and variance, [tex]\sigma^{2}_{s}=n\sigma^{2}[/tex].

Compute the probability of T < 12 as follows:

[tex]P(T<12)=P(\frac{T-\mu_{T}}{\sqrt{\sigma_{T}^{2}}}<\frac{12-(70\times0.20)}{\sqrt{70\times 0.04}})\\=P(Z<-1.19)\\\=1-P(Z<1.19)\\=1-0.883\\=0.117[/tex]

*Use a z-table for the probability.

Thus, the probability that all 70 messages are transmitted in less than 12 minutes is 0.117.

Following are the solution to the given question:

Let X signify the letter's transmission time  [tex]i^{th}[/tex] , and [tex]i =1,2,3,...,70.[/tex] this is assumed that the  [tex]X_i \sim Exp (\lambda =5)[/tex].

[tex]\to Mean =\frac{1}{\lambda} =\frac{1}{5}=0.2\\\\ \to Variance =\frac{1}{\lambda^2} =\frac{1}{5^2} =\frac{1}{25}=0.04\\\\[/tex]

               mean [tex]= E(T) = 70 \times 0.2=14[/tex]  

               Variance [tex]=V(T) =70 \times 0.04 = 2.8[/tex]  

       Therefore

             [tex]\to P(T<12) = P (Z < \frac{12-14}{\sqrt{2.8}})[/tex]

                                    [tex]= P(Z<-1.195) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (by sy memetry) \\\\=0.5-P(0<Z<1.19)\\\\=0.5-0.3830 \\\\=0.1170\\\\[/tex]

Therefore, the final answer is "0.1170".

Learn more about the Central Limit Theorem:

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Answer:

The correct answers are

option(1), option (4),option (5)

Step-by-step explanation:

One to one: Every image has exactly one unique pre-image in domain.

(1)

f(a) = b, f(b)=a, f(c)=c, f(d)=d

b has a pre-image in domain i.e a

a has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is one to one.

(2)

f(a) = b, f(b)=a, f(c)=c, f(d)=d

b has a pre-image in domain i.e a

a has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is one to one.

(3)

f(a) = b, f(b)=b, f(c)=d, f(d)=c

b has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

b has two pre image.

Here all elements have not a unique pre- image in domain.

Therefore it is not a one to one mapping.

(4)

f(a) = b, f(b)=b, f(c)=d, f(d)=c

b has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

b has two pre image.

Here all elements have not a unique pre- image in domain.

Therefore it is not a one to one mapping.

(5)

f(a) = d, f(b)=b, f(c)=d, f(d)=c

d has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is a one to one mapping.

(6)

f(a)=d, f(b)=b,f(c)=c,f(d)=d

d has a pre-image in domain i.e a

b has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is a one to one mapping.

f(a) = b, f(b) = a, f(c) = c, f(d) = d and f(a) = d, f(b) = b, f(c) = c, f(d) = d are one-to-one function and f(a) = b, f(b) = b, f(c) = d, f(d) = c is not one-to-one function. Then the correct statements are 1, 4, and 5.

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

One-to-one - every image has exactly one unique pre-image in the domain.

1. f(a) = b, f(b) = a, f(c) = c, f(d) = d

b has a pre image in a domian that is a

a has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

2. f(a) = b, f(b) = a, f(c) = c, f(d) = d

b has a pre image in a domian that is a

a has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

3. f(a) = b, f(b) = b, f(c) = d, f(d) = c

b has a pre image in a domian that is a

b has a pre image in a domian that is b

d has a pre image in a domian that is c

c has a pre image in a domian that is d

Here all elements do have not a unique pre-image in a domain.

Therefore it is not one-to-one.

4. f(a) = b, f(b) = b, f(c) = d, f(d) = c

b has a pre image in a domian that is a

b has a pre image in a domian that is b

d has a pre image in a domian that is c

c has a pre image in a domian that is d

Here all elements do have not a unique pre-image in a domain.

Therefore it is not one-to-one.

5. f(a) = d, f(b) = b, f(c) = c, f(d) = d

d has a pre image in a domian that is a

b has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

6. f(a) = d, f(b) = b, f(c) = c, f(d) = d

d has a pre image in a domian that is a

b has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

More about the function link is given below.

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Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.

The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.

In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

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Answer:

0.209

Step-by-step explanation:

Find the sample mean and standard deviation.

μ = 1050

s = 218 / √50 = 30.8

Find the z-score.

z = (1075 − 1050) / 30.8

z = 0.81

Find the probability.

P(Z > 0.81) = 1 − 0.7910 = 0.2090

Answer:

y= 65 degrees. and x=30 degrees

Step-by-step explanation:

i haven't done this for a while so i may or may not be correct i tried to help though :)

Answer: y = 52°. x = 64°

Step-by-step explanation:

y = 26 + 26 = 52 degrees ( exterior angle of a triangle is the sum of the two opposite interior angle of the triangle)

x = 180-(52+64)(sum of angles in a triangle is 180°)

x = 180-116

x = 64degrees ( base angles of an isosceles triangle are equal)

The maximum height reached by the shell is approximately 18255.79 meters.

To find when the shell reaches its maximum height and the value of the maximum height, we need to consider the forces acting on the shell and analyze its motion.

1. Force due to gravity:

The force due to gravity is given by the formula:

Force_gravity = mass × acceleration_due_to_gravity

Here, the mass of the shell is 2 kg, and the acceleration due to gravity is 9.81 m/s².

2. Force due to air resistance:

The force due to air resistance is given by the formula:

Force_air_resistance = (0.1) × velocity²

Here, the velocity of the shell is given as 600 m/s.

Using Newton's second law, we can calculate the net force acting on the shell:

Net force = Force_gravity - Force_air_resistance

When the shell reaches its maximum height, the net force is equal to zero because there is no acceleration at that point. Therefore, we can set the net force equation to zero and solve for the time:

0 = Force_gravity - Force_air_resistance

mass × acceleration_due_to_gravity = (0.1) × velocity²

2 kg × 9.81 m/s² = (0.1) × (600 m/s)²

Simplifying further, we find:

19.62 = 0.1 × 360,000

Time = 600 m/s / 9.81 m/s²

Time ≈ 61.15 seconds

Therefore, the shell will reach its maximum height approximately 61.15 seconds after being shot upward.

To find the maximum height, we can use the kinematic equation:

h = v₀t - (1/2)gt²

Substituting the given values into the equation, we find:

h = (600 m/s) × (61.15 s) - (1/2) × (9.81 m/s²) × (61.15 s)²

h ≈ 18255.79 meters

Therefore, the maximum height reached by the shell is approximately 18255.79 meters.

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The shell will reach its maximum height after approximately 4.51 seconds and this maximum height is around 100 meters.

This problem involves the physics of motion under the influence of gravity and air resistance. An important concept here is the 'terminal velocity', which is the velocity at which the upward force (due to the initial impulse provided to the shell) balances out the downward force (due to gravity and air resistance).

First, we must establish that the terminal velocity 'v' of the shell upwards will be achieved when the sum of the forces acting on it are zero. This gives us:

0 = -0.1v^2 + 2 * 9.81

. Solving this quadratic equation reveals v = sqrt(2*9.81/0.1) m/s ≈ 44.3 m/s.

After achieving this terminal velocity, the shell will start decelerating at a rate of 9.81 m/s^2 (the gravity acceleration). The time 't' that takes for the shell to stop moving upwards (so to reach its maximum height) can be calculated using the formula:

t = v / gravity = 44.3/9.81 ≈ 4.51 seconds

.

As for the maximum height 'h', it can be calculated using this formula:

h = v * t + 0.5 * (-gravity) * t^2

By inserting the values of v, gravity, and t in this equation, we find h ≈ 100 m.

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Answer:

Null hypothesis: [tex]p \geq 0.7[/tex]

Alternative hypothesis: [tex]p<0.7[/tex]

A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.

So the correct option for this case would be:

c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.  

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.  

Solution to the problem

On this case we want to test if the sporting goods store claims that at least 70^ of its customers, so the system of hypothesis would be:

Null hypothesis: [tex]p \geq 0.7[/tex]

Alternative hypothesis: [tex]p<0.7[/tex]

A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.

So the correct option for this case would be:

c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.

Answer:

Step-by-step explanation:

A. Given

  T = 0.05x +35 . . . . Terri's cost of operating a craft booth

  D = 0.03x +55 . . . . Donna's cost of operating a craft boot

  T = D

where x is the dollar amount of sales.

__

B. Solution

Subtracting the equation for D from that of T, we get ...

  T - D = 0

  (0.05x +35) -(0.03x +55) = 0 = 0.02x -20

  0 = x -1000 . . . . . divide by 0.02

  x = 1000

  T = D = 0.05(1000) +35 = 85

  (x, T) = (x, D) = (1000, 85)

__

C. Meaning

According to the given definitions of the variables, each booth pays a total of $85 in fees for sales of $1000.

The system of equations is T = 0.05x + 35 and D = 0.03x + 55. The solution is (1000, 85), meaning Terri and Donna each earned $1000 from sales and paid $85 total in booth and commission fees.

The system of equations to model the amount of money Terri and Donna pay each weekend at the craft shows can be written as follows: T = 0.05x + 35 and D = 0.03x + 55.

Since we know that Terri and Donna both paid the same total amount of money for their booths and commission, it means T = D. Or, we can equate the two equations: 0.05x + 35 = 0.03x + 55. Solving for x gives us x = 1000. Substituting x = 1000 into T = 0.05x + 35 equation, we get T (and D) = 85. So, the solution to the system of equations is (1000, 85).

In the context of this situation, the solution means that Terri and Donna both earned $1000 from sales, and each paid $85 total for their booth and commission fees.

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There is a useful linear relationship between rainfall and runoff and the relationship is y = 0.85x - 2.65

Deciding whether there is a useful linear relationship between rainfall and runoff

From the question, we have the following parameters that can be used in our computation:

x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127

y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105

Using a graphing tool, we have the following summary

The regression equation can be represented as

y = bx + a

Where

b = SP/SSX = 17313.93/20086.93 = 0.86

a = MY - bMX = 43.27 - (0.86*53.27) = -2.65

So, we have

y = 0.85x - 2.65

Hence, there is a useful linear relationship between rainfall and runoff

Answer:

a) 0.00563

b) 1

c) 0

Step-by-step explanation:

Total = 40+30+20 =90

a) (20C8×30C6×40C6)/90C20

= 0.00563

b) 1 - (no chocolate + 1 chocolate)

1 - [(50C20) + (40C1×50C19)]/90C20

1 - 0.00002478

= 0.9999752187

c) [40C20+20C20+30C20]/90C20

= 0.0000000027045

This is about permutations and combinations.

a) Probability = 0.00563

b) Probability = 0.99997522

c) Probability = 0.0000000027045

Chocolate = 40

Coconut = 30

Banana = 20

Total number of chocolates = 40 + 30 + 20

Total number of chocolates = 90

a) Probability that she will choose 8 banana and 6 chocolate cakes if she chooses 20 cupcakes at random will be;

(20C₈ × 30C₆ × 40C₆)/90C₂₀

(125970 × 593775 × 3838380)/50980740277700939310

This gives us   0.00563

b) Probability of at least 2 chocolate cupcakes is;

1 - [P(no chocolate) + P(1 chocolate)]

P(no chocolate) = (50C₂₀)/90C₂₀

P(1 chocolate) = (40C₁ × 50C₁₉)/90C₂₀

Thus;

1 - [P(no chocolate) + P(1 chocolate)] = 1 - [(40C₁ × 50C₁₉) + 50C₂₀]/90C₂₀

This gives us;  0.99997522

c) Probability of getting all cupcakes of same kind is;

(40C₂₀ + 20C₂₀ + 30C₂₀)/90C₂₀

⇒ 0.0000000027045

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Answer:

Franklin Freightways experienced $66 million.

Step-by-step explanation:

= ( Pretax accounting income + Overweight fines - Temporary difference: Depreciation) * tax rate

($200 + 5 - $40) × 40%

=Tax liability of $66.

Answer:

33% probability that the company will win at least one of the two contracts

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a company is awarded the first contract.

B is the probability that a company is awarded the second contract.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that a company is awarded the first contract but not the second and [tex]A \cap B[/tex] is the probability that a company is awarded both contract.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

The probability that it will get both contracts is .13.

This means that [tex]A \cap B = 0.13[/tex]

The probability that it will be awarded a second contract is .21

This means that [tex]B = 0.21[/tex]

[tex]B = b + (A \cap B)[/tex]

[tex]0.21 = b + 0.13[/tex]

[tex]b = 0.08[/tex]

The probability that a company will be awarded a certain contract is .25

This means that [tex]A = 0.25[/tex]

[tex]A = a + (A \cap B)[/tex]

[tex]0.25 = a + 0.13[/tex]

[tex]a = 0.12[/tex]

What is the probability that the company will win at least one of the two contracts?

[tex]A \cup B = a + b + A \cap B = 0.12 + 0.08 + 0.13 = 0.33[/tex]

33% probability that the company will win at least one of the two contracts

The probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\) or 0.34.[/tex]

To find the probability that the company will win at least one of the two contracts, we can use the principle of inclusion-exclusion. The principle states that the probability of the union of two events (in this case, winning either contract) is the sum of the probabilities of each event occurring individually, minus the probability of both events occurring together.

Let [tex]\(P(A)\)[/tex] be the probability that the company will win the first contract, [tex]\(P(B)\)[/tex] be the probability that the company will win the second contract, and [tex]\(P(A \cap B)\)[/tex] be the probability that the company will win both contracts. We are given:

[tex]\(P(A) = 0.25\), \(P(B) = 0.21\), \(P(A \cap B) = 0.13\).[/tex]

The probability that the company will win at least one contract, [tex]\(P(A \cup B)\)[/tex], is given by:

[tex]\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\][/tex]

Substituting the given probabilities:

[tex]\[P(A \cup B) = 0.25 + 0.21 - 0.13\] \[P(A \cup B) = 0.46 - 0.13\] \[P(A \cup B) = 0.33\][/tex]

To express this probability as a fraction, we can write 0.33 as [tex]\(\frac{33}{100}\)[/tex], which simplifies to[tex]\(\frac{17}{50}\)[/tex] when reduced to its simplest form.

Therefore, the probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\)[/tex] or 0.34

Using the limit laws of addition and multiplication, the limit of the given equation is solved by combining the calculated limits of the function and the constant, giving an answer of approximately 13.78.

To solve this problem, we can use the limit laws of addition and multiplication. The limit laws state that the limit of the sum of two functions is equal to the sum of their individual limits, and the limit of the product of two functions is equal to the product of their individual limits.

So, based on these laws, we first separate the function into two parts: the limit of 3√(f(x).g(x)) as x approaches 3 and the limit of 10 as x approaches 3.

Given lim f(x)=>6 and lim g(x)=>9 when x approaches 3, we multiply these values together to get a product of 54. The cubed root of 54 is approximately 3.78.

The constant 10 has a limit of 10, as constants maintain their value irrespective of the limit.

Adding these values together, 3.78 + 10, gives us a limit of approximately 13.78.

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Answer:

The parking rate at the hospital a parking garage are 50 cents for the 1st hour and 25 cents for each additional hour if guan part in the hospital parking garage for 8 hours how much will the total of charge for parking be

mADC=360-mABC=360-184=176⁰

So, measure of angle ABC= mADC/2=176/2=88⁰

The required measure of angle ABC is 88° for the given figure. The correct answer is option A.

The arc is a portion of the circumference of a circle. The circumference of a circle is the distance or perimeter around a circle

The measure of the arc is given as follows:

mABC = 184°

According to the given figure, mADC is the part of the full circle which is complete arc that measure of angle is 360°.

mADC = 360° - mABC

mADC = 360° - 184°

mADC = 176°

As we know that the measure of angle ABC is equal to half of mADC.

The measure of angle ABC = mADC/2

The measure of angle ABC = 176/2

The measure of angle ABC = 88°

Therefore, the required measure of angle ABC is 88°.

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Answer:

poalo is correct

Step-by-step explanation

Answer:

Neither Marco nor Paolo is correct, because neither one used the rational exponent property correctly.

Step-by-step explanation:

I just did the test and it told me that I got it right.

Hope that helps! Please mark me brainliest.

The probability that none of their four children has the disease = 0.316

Step-by-step explanation:

Step 1 :

The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25

So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75

Step 2 :

There are four children. The probability that each child has the disease is independent of the condition of the other children

The probability of more than one independent events happening together can be obtained by multiplying  probability of individual events

So, the probability that none of their 4 children has the sickle cell disease  = [tex](0.75)^{4}[/tex] = 0.316

Step 3 :

Answer :

The probability no children has the specified cell disease = 0.316