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kati45 [8]
2 years ago
12

Jeremiah can run 5 miles in 37 minutes. At that rate, how long does it take him to run 1 mile?

Mathematics
1 answer:
VashaNatasha [74]2 years ago
3 0
So what you would do is divide 37 minutes by 5 to get 7.4 minutes

answer: 7.4 minutes
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54 divided by 2,808 the answer an worked out
NemiM [27]
0.019 is the correct answer. I hope this is the answer you are looking for! :)
4 0
3 years ago
The Mad Hat Company must ship two different-sized boxes: The small ones cost them 45 cents each and weigh 6 ounces, and the larg
Viefleur [7K]

Given :

The Mad Hat Company must ship two different-sized boxes: The small ones cost them 45 cents each and weigh 6 ounces, and the large ones weigh 25 ounces and cost $1.20 each.

The total shipment this morning weighed 20 pounds 7 ounces and cost $18.45.

To Find :

How many packages of each size were shipped.

Solution :

We know, 1 pound = 16 ounces.

So, 20 pound = 20×16 = 320 ounces.

Let, number of large and small box are l and s.

So,

25l + 6s = 327 ...1)

0.45s + 1.20l = 18.45 ...2)

Solving both the equations, we get :

x = 9 and y = 17

Therefore, package smaller and big size are 17 and 9 respectively .

Hence, this is the required solution.

6 0
3 years ago
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

6 0
3 years ago
What is the sales tax of a $178.90 video game system is the tax rate is 3.75%
juin [17]
It would be: 178.90 * 3.75%
= 178.90 * 0.0375
= 6.71

In short, Sales tax would be $6.71

Hope this helps!
3 0
3 years ago
PLZ HELP I DONT UNDERSTAND A password for a security system must consist of a eater followed by two digits. How many possible pa
Talja [164]

Answer:

7020

Step-by-step explanation:

The letter could be any 1 of the 26 and it could be placed in any 1 of 3 places so this makes 26*3 = 78 possibilities.

For each of these  there is any one of  permutations of 2 digits from 10

= 10P2 = 10! / 8! = 10*9 = 90.

So the answer is 78 * 90 =  7020.

Note I am assuming that the 2 digits picked are different - no duplicates.

please mark me brainliest!

4 0
3 years ago
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