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stepladder [879]
3 years ago
15

A fruit company delivers its fruit in two types of boxes: large and small. a delivery of 6 large boxes and 5 small boxes has a t

otal weight of 127 kilograms. a delivery of 2 large boxes and 3 small boxes has a total weight of 51 kilograms. how much does each type of box weigh?
Mathematics
1 answer:
Afina-wow [57]3 years ago
5 0
6 large + 5 small = 127 kg ------------ (1)
2 large + 3 small = 51kg    ------------ (2)

<u>(2) x 3 :</u>
6 large + 9 small = 153 ---------------- (2a)

<u>(2a) - (1) :</u>
4 small = 26
1 small = 6.5kg   ------------ (sub into equation 1)

6 large + 5 small = 127 kg
6 large + 5(6.5) = 127
6 large + 32.5 = 127
6 large = 127 - 32.5
6 large = 94.5
1 large = 15.75 kg

Answer: The small box weighs 6.5kg and the big box weighs 15.75 kg


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15 units.

Step-by-step explanation:

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So here it is:

√(10- -2)^2 + (6- -3)^2)

= √(144+81)

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Find the value of k that makes the slope between the points (k,3) and (1,-4) equal to 0.75
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A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statisti
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We need a sample of size at least 13.

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

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

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95% confidence level

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Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

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

0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}

0.08\sqrt{n} = 1.96\sqrt{0.54*0.46}

\sqrt{n} = \frac{1.96\sqrt{0.54*0.46}}{0.08}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.54*0.46}}{0.08})^{2}

n = 12.21

Rounding up

We need a sample of size at least 13.

3 0
3 years ago
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