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choli [55]
3 years ago
15

Evaluate 21 - n for n = 3 * please help me with this

Mathematics
2 answers:
Alinara [238K]3 years ago
8 0
21-3=18 because n=3 and you’re just subtracting
qwelly [4]3 years ago
4 0

Answer:

18

Step-by-step explanation:

21-n; n=3

Substitute n for 3:

  • 21-3

Solve.

  • 18
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An apple pie uses 4 cups of apples and 3 cups of flour. An apple cobbler uses 2 cups of apples and 3 cups of flour. You have 16
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Part 1:

Given that x represents the <span>number of apple pies that can be made and y represents the number of apple cobblers that can be made.

Given that a</span><span>n apple pie uses 4 cups of apples and an apple cobbler uses 2 cups of apples and there are 16 cups of apples available.

</span>A<span>n inequality to show the constraint on the amount of apples available is given by:

4x + 2y ≤ 16
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Part 2:

<span>Given that an apple pie uses 3 cups of flour and an apple cobbler uses 3 cups of flour and there are 15 cups of flour available.

</span>A<span>n inequality to show the constraint on the amount of flour available is given by:

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Part C:

The </span><span>non negativity contraints on x and y are:

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3 years ago
Evaluate the expression<br> 7^0
Marysya12 [62]

Answer:

1

Step-by-step explanation:

Any number put to the 0 power is equal to 1.

Does that help?

3 0
3 years ago
What represents 200 million<br>​
sattari [20]

Answer:

200 million represents a very large number and therefore it is not commonly encountered in the history of the people of God. However, is reflected in the number of silver coins annually paid by the vinedressers who farmed King Solomon's vineyards (ie, 1,000 x 1,000 x 200).

4 0
3 years ago
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The
Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that n = 387, \pi = \frac{298}{387} = 0.77

84% confidence level

So \alpha = 0.16, z is the value of Z that has a p-value of 1 - \frac{0.16}{2} = 0.92, so Z = 1.405.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

6 0
3 years ago
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