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Vesnalui [34]
3 years ago
12

Shawn's Breakfast Goodies recently sold 3 blueberry muffins and 3 other muffins. What is the experimental probability that the n

ext muffin sold will be a blueberry muffin?
Mathematics
1 answer:
katovenus [111]3 years ago
3 0

Answer:

The experimental probability can be expressed in three ways:

Fraction Form: 3/6 or 1/2

Decimal Form: 0.5

Percent Form: 50%

Step-by-step explanation:

Experimental probability is expressed as:

<em>(Favorable outcomes) / (Total outcomes)</em>

The favorable outcome in this situation is blubbery muffins. There were 3 blueberry muffins sold, so this is the <u>numerator</u> of our fraction.

To find the total outcomes we add 3 + 3, which equals 6. This will be the <u>denominator</u> of our fraction.

The fraction used to express experimental probability will look like this: 3/6

We can further simplify this to 1/2.

3 ÷ 3 = 1

6 ÷ 3 = 2

To find a decimal, we divide 1 ÷ 2 = 0.5

Multiply this by 100 to get a percentage, 0.5 x 100 = 50%

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8 0
3 years ago
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Colin buys a car for £28100.
andriy [413]

Answer:

The amount the car will be worth after 6 years is £21,917 and 22 pence

Step-by-step explanation:

The amount for which Colin buys the car, PV = £28,100

The amount by which the car depreciates each year, r = 4%

The number of years after which the value of the car is sought, n = 6 years

The future value, FV, based on an annual depreciation is given as follows;

FV = PV \times \left ( 1 + \dfrac{r}{100} \right ) ^n

Substituting the known values gives;

FV_6 = 28,000 \times \left ( 1 - \dfrac{4}{100} \right ) ^6 =  21,917.22

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5 0
3 years ago
A lawn service owner is testing new weed killers. He discovers that a particular weed killer is effective 89% of the time. Suppo
marissa [1.9K]

Answer:

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564.

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}.

He discovers that a particular weed killer is effective 89% of the time. Suppose that this estimate was based on a random sample of 60 applications.

This means that \pi = 0.89, n = 60

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.89 - 1.645\sqrt{\frac{0.89*0.11}{60}} = 0.8236

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.89 + 1.645\sqrt{\frac{0.89*0.11}{60}} = 0.9564

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564.

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A=9

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