x y

serious [3.7K]

For the #2 it would be 3 because 50 goes into 138 2 times which will give you 100 with a remainder of 38. You can't leave behind the other 38 so you will bring another bus.

4 0

3 years ago

A moving sidewalk in an airport moves people between gates. It takes Jason's 8-year-old

likoan [24]

Answer:

The sidewalk moves at 0.5 ft/sec

Josie's speed walking on a non-moving ground is 3.5ft/sec

Step-by-step explanation:

Let x represent the speed of the side walk and y represent her walking speed

It takes Jason's 8-year-old daughter Josie 44 sec to travel 176 ft walking with the sidewalk

Distance = speed × time

176 = (x+y)×44

44x+44y = 176

x+y = 4 .......1

It takes her 7 sec to walk 21 ft against the moving sidewalk in the opposite direction).

21 = (y-x)7

7y - 7x = 21

y - x = 3 ......2

Add equation 1 to 2

2y = 7

y = 3.5 ft/sec

From equation 1

x + y = 4

x = 4 - 3.5 = 0.5

x = 0.5 ft/sec

The sidewalk moves at 0.5 ft/sec

Josie's speed walking on a non-moving ground is 3.5ft/sec

6 0

3 years ago

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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!}
$

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!}
$

$\simeq 0.7408182207$ ------------------------------------(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!}
$

$\simeq 0.2222454662$ ------------------------------------(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)

7 0

3 years ago

The measure of angle M is_degrees.

jasenka [17]

Answer:

m=37

Step-by-step explanation:

The 2 angles are complementary, or add to 90 degrees. Wee know this because of the little square in the corner that indicates a right angle, which is equal to 90 degrees.

m+53=90

Solve for m by getting m by itself. To do this, subtract 53 from both sides to "reverse" the addition being done to m.

m+53-53=90-53

The 53s on the left cancel, and we are left with:

m=37

3 0

3 years ago

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Two mechanics work on a car . The first mechanic worked 10 hours and the second mechanic worked for 15 hours. Together they char

bonufazy [111]

Answer:

110

Step-by-step explanation:

Let x = rate #1 and then 155-x=rate #2.

15x + 5(155-x) = 1225

15x - 5x + 775 - 1225

10x = 450

x = 45, second rate is 110

8 0

3 years ago

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