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vichka [17]
3 years ago
13

A rowing team rowed 90 miles while going with the current in the same amount of time as it took to row 10 miles going against th

e current. The rate of the current was 4 miles per hour. Find the rate of the rowing team in still water.
Mathematics
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:   
 let x=rate of the rowing team in still water 
 x+4=rate of the rowing team with current 
 x-4=rate of the rowing team against current   
 travel time=distance/rate 
 .. 
 90%2F%28x%2B4%29=10%2F%28x-4%29 
 90x-360=10x+40 
 80x=400 
 x=5  rate of the rowing team in still water=5 mph

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ANSWER: B

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Perimeter = L + L + W + W

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My professor showed us in class today how to use ode45 to solve a differential equation numerically. i would like to use it on m
Brilliant_brown [7]

Yes, ode45 can be used for higher-order differential equations. You need to convert the higher order equation to a system of first-order equations, then use ode45 on that system.

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7 0
3 years ago
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kobusy [5.1K]

Answer:

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8 0
3 years ago
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A bank manager has developed a new system to reduce the time customers spend waiting for teller service during peak hours. The m
Lana71 [14]

Answer:

(a) <em>H₀</em>: <em>μ</em> = 10 vs. <em>Hₐ</em>: <em>μ</em> < 10.

(b) The level of significance is 0.05.

Step-by-step explanation:

A new system is used to reduce the time customers spend waiting for teller service during peak hours at a bank.

A single mean test can be used to determine whether the waiting time has reduced.

(a)

The hypothesis to test whether the new system is effective or not is:

<em>H₀</em>: The mean waiting time is 10 minutes, i.e. <em>μ</em> = 10.

<em>Hₐ</em>: The mean waiting time is less than 10 minutes, i.e. <em>μ</em> < 10.

(b)

The information provided is:

\bar x=9.5\\s=2.2\\n=70

Compute the test statistic value as follows:

t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{9.5-10}{2.2/\sqrt{70}}=-1.902

The test statistic value is <em>t</em> = -1.902.

Compute the <em>p</em>-value of the test as follows:

p-value=P(t_{n-1}

               =P(t_{69}1.902)\\=0.031

The null hypothesis will be rejected if the <em>p</em>-value of the test is less than the significance level (<em>α</em>).

The <em>p</em>-value obtained is 0.031.

To reject the null hypothesis the value of <em>α</em> should be more than 0.031.

The most commonly used values of <em>α</em> are: 0.01, 0.05 and 0.10.

So, the least value of <em>α</em> at which we can conclude that the wait times have decreased is, <em>α</em> = 0.05.

Thus, the level of significance is 0.05.

6 0
3 years ago
HELP! WILL AWARD BRAINLIEST TO WHOEVER ANSWERS BOTH PARTS CORRECTLY!!
rewona [7]

check the picture below.

now, we're assuming the trapezoid is an isosceles trapezoid, namely AD = BC, and therefore the triangles are twins.

incidentally, b is the height of the trapezoid and likewise is also the altitude or height of the concrete triangle.

so we can simply get the area o the trapezoid, notice the bottom base is a+185+a, and then get the area of the concrete triangle and subtract the triangle from the trapezoid, what's leftover is just the vegetation area.

\bf \begin{cases} a=283\cdot cos(80^o)\\ a\approx 49.14\\ --------\\ b=283\cdot sin(80^o)\\ b\approx 278.70 \end{cases}\\\\ -------------------------------\\\\ \textit{area of a trapezoid}\\\\ A=\cfrac{h(x+y)}{2}~~ \begin{cases} x,y=\stackrel{bases}{parallel~sides}\\ h=height\\ ----------\\ x=185\\ y\approx \stackrel{a+185+a}{283.28}\\ h\approx\stackrel{b}{278.70} \end{cases} \\\\\\ A=\cfrac{278.70(185+283.28)}{2}\implies A\approx 65254.818

so that's the area of the trapezoid, now let's get the area of the triangle.

\bf \stackrel{triangle}{\cfrac{1}{2}(185)(b)}\implies \cfrac{1}{2}(185)(278.70)\qquad \approx 25779.80\\\\ -------------------------------\\\\ \stackrel{\textit{area for vegetation}}{\stackrel{\textit{area of trapezoid}}{65254.818}~~-~~\stackrel{\textit{area of triangle}}{25779.80}}\implies 39475.018

since we know 36 yd² cost 12 bucks, then how much will it be for 39475.018 yd²?

\bf \begin{array}{ccll} yd^2&\$\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 36&12\\ 39475.018&x \end{array}\implies \cfrac{36}{39475.018}=\cfrac{12}{x}\implies x=\cfrac{39475.018\cdot 12}{36} \\\\\\ x\approx 13158.339\overline{3}

3 0
3 years ago
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