What is the value of x

Inez Walks 1.85 mi in 45 minutes , how many miles can she walk in 1 hour?

insens350 [35]

Answer:

She can walk 2.46 repeating miles in 1 hour

Step-by-step explanation:

We can write a proportion to solve this problem. Write the number of miles on top and the time in minutes on the bottom. We know 1 hour is 60 minutes.

1.85 miles x miles

------------------- = ----------------------

45 miles 60 minutes

Using cross products

1.85 * 60 = 45x

Divide each side by 45

1.85*60/45 = 45x/45

2.466666666 =x

She can walk 2.46 repeating miles in 1 hour

4 0

3 years ago

How many times does 31 go into 248?

valina [46]

248 / 31 = 217 times

5 0

3 years ago

Elena’s bedroom door is 0.8 m wide. How wide should the door be on the scale drawing, if the ratio is 1 to 50?

arlik [135]

We are given Elena’s bedroom door's width = 0.8 m.

Also the scale drawing is in the ratio of 1 to 50 that is 1/50.

<em>In order to find the width of scale drawing, we need to multiply original width of the door by 1/50.</em>

If we multiply 0.8 by 1/50, we get

0.8 × 1/50 = 0.8/50 = 0.016 meter.

So, we can say 0.016 meter wide should the door be on the scale drawing, if the ratio is 1 to 50.

7 0

4 years ago

Read 2 more answers

On a map with a scale 1 : 100,000, the distance between two cities is 12 cm. What would be the distance between these two cities

miskamm [114]

Answer:

36 cm

Step-by-step explanation:

5 0

1 year ago

The number of people arriving for treatment at an emergency room can be modeled by aPoisson process with a rate parameter of 5/h

saveliy_v [14]

Answer:

(a) The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is 3.75 arrivals.

Step-by-step explanation:

(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:

$P(X=4)=\frac{\lambda^{X}*e^{-\lambda} }{X!} =\frac{5^{4}*e^{-5} }{4!}=\frac{625*0.006737947}{24} =\frac{4.211}{24}=0.1754$

The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour can be written as

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))$

Using

$P(X)=\frac{\lambda^{X}*e^{-\lambda} }{X!}$

We get

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))\\P(X>3)=1-(0.0067+ 0.0337+ 0.0842 + 0.1404 )\\P(X>3)=1-0.2650=0.7350\\$

The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is

$EV=\lambda*t=5*0.75=3.75$

8 0

3 years ago