Bill and Will, starting together, ran a 400-meter race, each running at a constant speed. When Bill crossed the finish line, Will was exactly 20 yards behind Bill. They decide to run the race again, this time Bill starting 20 yards behind the original starting line and each running at his same constant speed as before. This time _______ wins by _______ yards.

Split up C into two component paths C₁ and C₂, where each line segment is respectively parameterized by

r₁(t) = (1 - t ) (3i + k) + t (4i + 3j + k) = (t + 3) i + 3t j + k

r₂(t) = (1 - t ) (4i + 3j + k) + t (4i + 5j + 4k) = 4i + (2t + 3) j + (3t + 1) k

both with 0 ≤ t ≤ 1.

It's a bit unclear what function you're supposed to integrate (looks like xyz ?) so I'll give a more general result. The line integral of a scalar function f(x, y, z) along the given path C is

[tex]\displaystyle \int_C f(x,y,z)\,\mathrm ds = \int_{C_1}f(\mathbf r_1(t))\left\|\frac{\mathrm d\mathbf r_1}{\mathrm dt}\right\|\,\mathrm dt + \int_{C_1}f(\mathbf r_2(t))\left\|\frac{\mathrm d\mathbf r_2}{\mathrm dt}\right\|\,\mathrm dt[/tex]

We have

dr₁/dt = i + 3j   ==>   || dr₁/dt || = √(1² + 3²) = √10

dr₂/dt = 2j + 3k   ==>   || dr₂/dt || = √(2² + 3²) = √13

Then the integrals reduce to

[tex]\displaystyle \int_0^1 \left(\sqrt{10}\,f(t+3,3t,1) + \sqrt{13}\,f(4,2t+3,3t+1)\right)\,\mathrm dt[/tex]

If indeed f(x, y, z) = xyz, then we have

[tex]\displaystyle \int_0^1 \left(3\sqrt{10}\,t(t+3) + 4\sqrt{13}\,(2t+3)(3t+1)\right)\,\mathrm dt = 11\sqrt{\frac52}+42\sqrt{13}[/tex]

Answer:

The range is the set of first coordinates

The selected statistical questions are as follows:

A what is the typical amount of rainfall for the month of June in the Galapagos islands

D how many songs does the class usually listen to each day

G how long does it typically take for 2nd graders to walk a lap around the track

Statistical questions are known by these two main characteristics:

This means that the data vary.  They are not the same data.  For example, "How old is Monday?" is not a statistical question because it has one answer.

Thus, statistical questions are not answered by a single number or answer, especially when the correct answer is just one.  

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a) the cost of a bunch of grapes compared with its weight

GRAAAAAAAAAAAAAAAAAAAAAAAAAAAAPES!!!!!

Answer:

A is the answer! I think you know because the formula is given at the top.

if the formula is [tex]\frac{(B+b)}{2} .h[/tex] we just need to plug in the values

21/2 = 10.5 x 5 = 52.5

hope it helps :)

Answer:

A = 52.5 in^2

Step-by-step explanation:

The area of a trapezoid is

A = 1/2 (b1+b2)h

where b1 and b2 are the bases and h is the height

A = 1/2 ( 14+7)*5

A = 105/2

A = 52.5 in^2

Answer:

answer is 10

explanation

when u add 7 with 10 u get 17 then double of 17 is 34

I hope It helps

It is found that the unknown number was 10.

An equation is an expression that shows the relationship between two or more numbers and variables. The addition is one of the mathematical operations. then the addition of two numbers results in the total amount of the combined value.

Given that "I add 7 to a certain number. I double the result. My final answer is 34".

Let consider the number be 10.

When we add 7 with 10 we get;

7 + 10 =  17

then double the result of 17 = 34

Hence, the unknown number was 10.

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#SPJ2

Answer:

y = -1/6x - 1.

Step-by-step explanation:

I  am assuming that the point id (6, -2).

The slope of the required line = -1/6.

y - y1 = m(x - x1) where m = slope and x1,y1 is a point on the line so we have

y - (-2) = -1/6( x- 6)

y + 2 = -1/6x + 1

y = -1/6x - 1.

Answer:

picture not clear what is the length

Answer:

answer is (4x^4+3y)(16x^8-12x^4y+9y^2)

Step-by-step explanation:

Answer:

-290

Step-by-step explanation:

(-260) +(-30)

=-260-30

=-290

Answer:

the answer is -290

Step-by-step explanation:

it is telling us to add so we can see that both the numbers have the same sign which is negative

when the signs are all the same, we can add and we got -290

Answer:

B. One of the populations is normally distributed.

Step-by-step explanation:

To test a claim about two population standard deviation or variance, it is imperative that the data meets certain requirements which include :

Randomness : Data must not be biased as such it must be drawn as a random sample from a larger group.

The data must be independent. That is not related to one another, the outcome of one should not rely on the outcome or value of another.

Both groups must be drawn From a population which is normally distributed.

One group being normally distributed by stribuyed while the other isn't a requirement for hypothesis testing in this scenario.

Answer: The pelican dove 5ft

Step-by-step explanation:

Pelican = 3ft

Fish = -2ft

Went down 3ft to get to 0, went down 2 more feet to get the fish.

Observe that the x coords of the roots of a polynomial are,

[tex]x_{1,2,3,4}=\{-3,0,1,4\}[/tex]

Which can be put into form,

[tex]y=a(x-x_1)(x-x_2)(x-x_3)(x-x_4)[/tex]

with data

[tex]y=a(x-(-3))(x-0)(x-1)(x-4)=ax(x+3)(x-1)(x-4)[/tex]

Now if I take any root point and insert it into the equation I won't be able to solve for y because they will always multiply to zero (ie. when I pick [tex]x=-3[/tex] the right hand side will multiply to zero,

[tex]y=-3a(-3+3)(-3-1)(-3-4)=0[/tex]

and a will be "lost" in the process.

If we observed a non-root point that we could substitute with x and y and result in a non-loss process then you could find a. But since there is no such point (I don't think you can read it of the graph) there is no other viable way to find a.

Hope this helps :)

[tex]\boxed{\sf a_n=\dfrac{(-1)^n}{5n}}[/tex]

a:-

[tex]\\ \sf\longmapsto a_1=\dfrac{(-1)^1}{5(1)}[/tex]

[tex]\\ \sf\longmapsto a_1=\dfrac{-1}{5}[/tex]

[tex]\\ \sf\longmapsto a_1=-5[/tex]

b:-

[tex]\\ \sf\longmapsto a_4=\dfrac{(-1)^4}{5(4)}[/tex]

[tex]\\ \sf\longmapsto a_4=\dfrac{1}{20}[/tex]

c:-

[tex]\\ \sf\longmapsto a_{30}=\dfrac{(-1)^{30}}{5(30)}[/tex]

[tex]\\ \sf\longmapsto a_{30}=\dfrac{1}{150}[/tex]

d:-

[tex]\\ \sf\longmapsto a_{19}=\dfrac{(-1)^{19}}{5(19)}[/tex]

[tex]\\ \sf\longmapsto a_{19}=\dfrac{-1}{95}[/tex]

Answer:

a) The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b) Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c) 37 cars would have to be sampled.

Step-by-step explanation:

Question a:

We have the sample standard deviation, and thus, the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.1448

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 2.1448\frac{6.2}{\sqrt{15}} = 3.4[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 26.7 - 3.4 = 23.3 mpg.

The upper end of the interval is the sample mean added to M. So it is 26.7 + 3.4 = 30.1 mpg.

The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b. What would happen to the interval if you increased the confidence level from 95% to 99%? Explain

Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c. The lead engineer is not happy with the interval you constructed and would like to keep the width of the whole interval to be less than 4 mpg wide. How many cars would you have to sample to create the interval the engineer is requesting?

Width is twice the margin of error, so a margin of error of 2 would be need. To solve this, we have to consider the population standard deviation as [tex]\sigma = 6.2[/tex], and then use the z-distribution.

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

How many cars would you have to sample to create the interval the engineer is requesting?

This is n for which M = 2. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]2 = 1.96\frac{6.2}{\sqrt{n}}[/tex]

[tex]2\sqrt{n} = 1.96*6.2[/tex]

[tex]\sqrt{n} = \frac{1.96*6.2}{2}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*6.2}{2})^2[/tex]

[tex]n = 36.9[/tex]

Rounding up:

37 cars would have to be sampled.

For engineering, it would be very useful to know Pythagorean theorem. You can use it to measure the tension in each ropes.

Answer:

B

Step-by-step explanation:

Answer:

D. 6.5

Step-by-step explanation:

The diameter of the cylinder is 1.25 x 2 = 2.5

SK = √1.25² + 6² = √42.25 = 6.5

Answer:

I am not so sure but I am assuming-5 2/3 to be mixed fraction so improper fraction wud be -17/3

now 4*- 17/3 = 68/3

now divide -5/8 ÷ 12

so u must know the rule that if division we can multiply by reciprocal

so -5/(8*12) = -5/96

hope that helps

Option D is correct. 95% of the distribution lies between 1.9975inches and 3.4025inches.

To get the required range of values, we will have to first get the z-score for the two-tailed probability at a 95% confidence interval. According to the normal table, the required range is between -2.81 and 2.81

The formula for calculating the z-score is expressed as;

[tex]z=\frac{x-\overline x}{s}[/tex] where:

[tex]\overline x[/tex] is the mean

s is the standard deviation

z is the z-scores

Given the following

[tex]\overline x[/tex]=2.7 in

s = 0.25

if z = -2.81

[tex]-2.81=\frac{x-2.7}{0.25}\\x-2.7=-2.81*0.25\\x-2.7=-0.7025\\x=-0.7025+2.7\\x=1.9975[/tex]

Similarly:

[tex]2.81=\frac{x_2-2.7}{0.25}\\x_2-2.7=2.81*0.25\\x_2-2.7=0.7025\\x_2=0.7025+2.7\\x_2=3.4025[/tex]

Hence the 95% of the distribution lies between 1.9975inches and 3.4025inches.

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Answer:

C

Step-by-step explanation:

Domain of the function is the whole R

Answer:

X=0,y=-5

x=4,y=-1

Step-by-step explanation:

Replace all occurrences of y with x^2-3x-5

(x^2-3x-5)+5=x

x^2-3x=x

X^2-4x=0

so :x=0,4

enter the value of x in the equation then find y

y=-5,-1

Answer:

6

Step-by-step explanation:

the | | are for absolute value, which means

|-6|=|6|= 6

Answer:

The rate at which the distance between the two cars is increasing is 30 mi/h

Step-by-step explanation:

Given;

speed of the first car, v₁ = 24 mi/h

speed of the second car, v₂ = 18 mi/h

Two hours later, the position of the cars is calculated as;

position of the first car, d₁ = 24 mi/h  x 2 h  = 48 mi

position of the second car, d₂ = 18 mi/h  x 2 h = 36 mi

The displacement of the two car is calculated as;

displacement, d² = 48² + 36²

                        d² = 3600

                        d = √3600

                        d = 60 mi

The rate at which this displacement is changing = (60 mi) / (2h)

                                                                                 = 30 mi/h

The area of a triangle is the product of its height and base divided by 2. The area of [tex]\triangle DEN[/tex] is [tex]555m^2[/tex]

Given that:

[tex]NE = 37[/tex]

[tex]DN = 36[/tex]

[tex]h = 30[/tex] --- altitude from point D

The altitude from point D will meet with side NE.

So, we have:

[tex]Base = NE = 37[/tex]

[tex]Height = 30[/tex]

The area of the triangle is:

[tex]Area = \frac{1}{2} \times base \times Height[/tex]

This gives:

[tex]Area = \frac{1}{2} \times 37 \times 30[/tex]

[tex]Area = 555[/tex]

Hence, the area of [tex]\triangle DEN[/tex] is [tex]555m^2[/tex]

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Answer:

B. 555 m²

Step-by-step explanation:

Answer:

6.8 feet

Answer From Gauth Math

Width = area/length = 22.6/5.2 = 113 ≈/26

Diagonal = √[length² + width²] ≈ √23.0 feet