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wolverine [178]
3 years ago
15

the Honda CRV gets 37 miles per gallon on average how much carbon dioxide will be admitted if the Honda were driven 400 miles

Mathematics
1 answer:
Pani-rosa [81]3 years ago
6 0

According to the equation located here: https://www.1728.org/glwarmng.htm burning one gallon of gasoline produces 19.3 pounds of carbon dioxide.

So, if a car gets 37 miles per gallon and if it travels 400 miles, it will use 10.81 gallons of gasoline and will produce 19.3 * 10.81 = 208.63 pounds of carbon dioxide.



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That’s a irrational number
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Mrs. Jasper has some 8-ounce cups for serving punch. Her punch bowl holds 2 gallons of punch. Mrs. Jasper thinks she can serve f
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She is wrong

Step-by-step explanation:

in one gallon there is 128 ounces, and 2 would make it 256.

40 - 8 ounce cups would make it 320 ounces, therefore she would need 3 gallons at least.

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Solve for pizza in the pie chart!
solniwko [45]

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Read 2 more answers
QUICK HELP!!!!
sasho [114]

Answer:

y =  \frac{3}{4} x -  \frac{3}{2}

Step-by-step explanation:

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- 6 \: from \: both \: sides

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divede \: by \: 2 \: on \: both \: sides

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7 0
3 years ago
Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was
andrezito [222]

Answer:

The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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For this problem, we have that:

n = 100, p = 0.42

92% confidence level

So \alpha = 0.08, z is the value of Z that has a pvalue of 1 - \frac{0.08}{2} = 0.96, so Z = 1.75.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.3336

The upper limit of this interval is:

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The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

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