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Monica [59]
3 years ago
10

The time it takes an excursion boat to travel 36 miles downstream is the same as the time it takes the boat to travel 12 miles u

pstream. If the stream is 4 miles per hour, which of the following is the speed of the boat in still water.
Mathematics
1 answer:
adoni [48]3 years ago
6 0

Hello there,

The time it takes an excursion boat to travel 36 miles downstream is the same as the time it takes the boat to travel 12 miles upstream. If the stream is 4 miles per hour, which of the following is the speed of the boat in still water.

Answer: 8 m/h

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What is the area of the rectangle shown on the coordinate plane? Enter your answer in the box. Do not round at any steps. units²
Goshia [24]
<h3>Answer: 42 square units</h3>

Check out the attached images below. In figure 1, I plotted the four points given as A, B, C, D; then connected them to form a rectangle.

Figure 2 shows 4 more points E, F, G, H made so that a larger rectangle forms. This newer larger rectangle entirely encloses the first rectangle. The idea is to find the area of the larger rectangle, and subtract off the areas of the four triangular pieces shown in figure 3. These images are attached below.

area of rectangle EFGH = base*height = 10*10 = 100

area of triangle ABE = base*height/2 = 3*3/2 = 4.5

area of triangle ADH = base*height/2 = 7*7/2 = 24.5

area of triangle CGD = base*height/2 = 3*3/2 = 4.5

area of triangle FCB = base*height/2 = 7*7/2 = 24.5

the four triangles have areas that add up to 4.5+24.5+4.5+24.5 = 58

Subtract this from the area of rectangle EFGH to get 100-58 = 42

----------

alternatively, you can find the distance from A to B getting roughly 4.24264 (use the distance formula). The distance from B to C is roughly 9.89949

Multiply those two values: 4.24264 * 9.89949 = 41.9999722536 which rounds to 42. There's rounding error based on the fact that the previously mentioned decimal values are approximate.

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4 years ago
1+1 lol dats da question
Gekata [30.6K]

Answer:

yellow?

Step-by-step explanation:

so you have 1 then add another an get yellow, bam im a genius

5 0
3 years ago
Tonia and trinny are twins. Their friends give them identical cakes for their birthday. Tonia eats 1/8 of her cake and trinny ea
Nikitich [7]

Answer:

\frac{7}{12} of the cake

Step-by-step explanation:

add \frac{1}{8} and \frac{1}{6} to see the total amount of cake eaten.

a. find the common denominator: 8 x 3 = 24 and 6 x 4 = 24

b. multiply accordingly to get the correct numerator: \frac{3}{24} + \frac{4}{24}

c. add: \frac{3}{24} + \frac{4}{24} = \frac{7}{24}

subtract found value from total to find left over cake.

a. 24 - 7 = 14

simplify.

a. \frac{14}{24} = \frac{7}{12}

You are left with \frac{7}{12} of the cake.

8 0
4 years ago
Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1
alexandr1967 [171]

Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, \frac{1.2}{4}

= 0.3 collisions per  month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}&#10;

                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}&#10;

                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}&#10;

                      \simeq 0.2222454662 ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

1 collision in 6 months period means

                                \frac{1}{6} collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

\simeq 0.6543894283-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = (P(X =0)^{6}

   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

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