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EleoNora [17]
2 years ago
8

A grocery store sells a bag of 6 oranges for $2.34. If Mav spent $1.95 on oranges, how many did she buy?

Mathematics
2 answers:
Yuri [45]2 years ago
8 0

Answer: She bought 5 oranges

Step-by-step explanation:

We need to find out how much one orange costs first, so we do:

$2.34 ÷ 6<u>or</u> = $0.39

Thirty-nine cents is how much one orange cost.

Now you multiply the initial value of 1 orange by 5, so:

$0.39 x 5 = $1.95, $1.95 =5 oranges

You can also do $1.95 divided by $0.39 which is 5.

(you can multiply till you find the number or divide the given number by the other divided number)

Therefore, that means Mav bought 5 oranges.

poizon [28]2 years ago
5 0

Answer:

5 oranges

Step-by-step explanation:

to solve this problem we need to make a cross-multiplication equation

2.34 is to 6 as 1.95 is to ?

? is the number of oranges that are needed to be found that were bought

2.34/6 = 1.95/?

Now we cross multiply 6 and 1.95 = 11.7

11.7/2.34 = 5

Give brainliest, please!
hope this helps :)

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Find the area of each figure. Round to the nearest hundredth where necessary.
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Answer:

The answer to your question is Triangle's area = 520 in², Square's area = 576 in²

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3 0
3 years ago
At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a
morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

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So, for exactly one car would be

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=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599

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We will find the traffic flow q such that

P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%

Hence, a) 4.6798, and b) 19.8%.

7 0
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