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Assoli18 [71]
3 years ago
5

On a trip to a lake, Kerrie and Shelly rode their bicycles four more than three times as many miles in the afternoon as in the m

orning. If the entire trip was 112 miles long, how far did they ride in the morning and how far in the afternoon?
Mathematics
2 answers:
iragen [17]3 years ago
6 0

Answer:

27 miles in the morning and 85 miles in the afternoon.

Explanation:

Let m be the number of miles they rode in the morning.  They rode 4 more than 3 times this many in the afternoon; this gives us the expression

3m+4

to represent the miles ridden in the afternoon.

Together they rode 112 miles; this means we add the morning miles, m, to the afternoon miles, 3m+4, and get 112:

m+3m+4 = 112

Combine like terms:

4m+4 = 112

Subtract 4 from each side:

4m+4-4= 112-4

4m = 108

Divide both sides by 4:

4m/4 = 108/4

m = 27

They rode 27 miles in the morning.

That means in the afternoon, they rode

3m+4 = 3(27)+4 = 81+4 = 85 miles.

gogolik [260]3 years ago
3 0
X+3x+4=112
x=27
3(27)+4=85

morning: 27 miles
afternoon: 85 miles
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Step-by-step explanation:

speed = distance/time

we know the speed and the time (11:45am to 1:45pm = 2 hours).

distance = speed × time = 56 m/h × 2 h = 112 miles

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What type of reaction is shown below? Check all that apply. 2Na + Cl2 → 2NaCl synthesis decomposition combustion
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Answer: This is a type of Synthesis reaction.

Explanation:

  • Synthesis reaction is a type of chemical reaction in which two or more substances combine to form a single substance.

A+B\rightarrow AB

  • Decomposition reactions are a type of chemical reactions in which a substance breaks down into two or more substances.

AB\rightarrow A+B

  • Combustion reaction is a type of chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water.

Hydrocarbon+O_2\rightarrow H_2O+CO_2

For the given reaction:

2Na(s)+Cl_2(g)\rightarrow 2NaCl

Here, 2 substances are combining together to form a single substance. Hence, it is considered as a synthesis reaction.

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Can somebody please help me 8th grade math puzzle
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3 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

Z = \frac{X - \mu}{\sigma}

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
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