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Usimov [2.4K]
3 years ago
10

Chen is bringing fruit and veggies to serve at an afternoon meeting. He spends a total of $28.70 on 5 pints of cut veggies and 7

pints of cut fruit. The food cost is modeled by the equation 5v+7f=28.70, where v represent the cost of one pint of cut veggies and f represents the cost of one pint of cut fruit. If the cost of each pint of fruits is $2.85, what is the approximate price of a pint of veggies
Mathematics
2 answers:
kobusy [5.1K]3 years ago
8 0
V=1.75

pls make me brainliest
alexgriva [62]3 years ago
7 0

Answer:

The the approximate price of a pint of veggies is $ 1.75.

Step-by-step explanation:

Here, v represents the cost of one pint of cut veggies and f represents the cost of one pint of cut fruit.

Also, the equation that models the given situation is,

5v + 7f = 28.70, ------(1)

Now, it is given that the cost of each pint of fruits is $2.85,

That is, f = $ 2.85,

Then, from equation (1),

5v + 7(2.85) = 28.70

5v + 19.95 = 28.70

5v = 8.75

v = 1.75,

Hence, the the approximate price of a pint of veggies is $ 1.75.

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2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for
kirza4 [7]

a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

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

The margin of error is given by:

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

The variables are listed as follows:

  • \pi is the sample proportion, which is also the estimate of the parameter.
  • z is the critical value.
  • n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of \frac{1+0.90}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

n = 150, \pi = \frac{15}{150} = 0.1

The margin of error is of:

M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%

The interval is given by the estimate plus/minus the margin of error, hence:

  • The lower bound is: 10 - 2.45 = 7.55%.
  • The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

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

0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}

\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}

(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

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