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1 year ago

6. The 4-inch radius pizza for $3 or the 8-inch radius pizza for $14.

Mathematics

1 answer:

SCORPION-xisa [38]1 year ago

8 0

Answer:

The 3 dollar pizza is the better deal.

Step-by-step explanation:

I take you mean which is the better deal.

The area of the 4 inch pizza is pi*r^2 = 3.14 * 4^2 = 50.24 square inches.

The area of the 8 inch pizza is pi*r^2 = 3.14 * 64 = 200.96

A square inch of the 4 inch pizza is 3/50.24 = 0.05952 cents

A square inch of the 8 inch pizza is 14/200.96 = 0.0697 cents.

Though this doesn't happen in real life, the smaller pizza is the better buy.

You might be interested in

The following table shows last year's revenue for each Herman's Hoagies location.

SOVA2 [1]

Answer:

The revenue for Granton location is 175 thousand dollars

Step-by-step explanation:

Given

Cedarton 121

Rimber 189

Linton 147

Mean = 158

Required

Revenue for Granton location.

To calculate the revenue for Granton location, we make use of mean formula.

Mean is calculated by Summation of Observation divided by number of observations.

Since Granton location is unknown; Let it be represented by letter G.

So, the summation of observation becomes 121 + 189 + 147 + G

Summation = 457 + G

The number of observations = 4

Recall that Mean = Summation ÷ Number

By substituting 158 for mean, 457 + G for summation and 4 for number, we have

158 = (457 + G) ÷ 4

158 = ¼(457 + G)

Multiply both sides by 4

4 * 158 = = 4 * ¼(457 + G)

632 = 457 + G

Make G the subject of formula

G = 632 - 457

G = 175

Hence, the revenue for Granton location is 175 thousand dollars

8 0

2 years ago

Read 2 more answers

A painter is painting a wall with an area of 150 ft2. He decides to paint half of the wall and then take a break. After his brea

Novosadov [1.4K]

The answer is 141.35 ft²

Before the first break, it was painted:
150 ft² ÷ 2 = 75 ft²
Now it's left:
150 ft² - 75 ft² = 75 ft²

Before the second break, it was painted:
75 ft² ÷ 2 = 37.5 ft²
Now it's left:
75 ft² - 37.5 ft² = 37.5 ft²

Before the third break, it was painted:
37.5 ft² ÷ 2 = 18.75 ft²
Now it's left:
37.5 ft² - 18.75 ft² = 18.75 ft²

Before the fourth break, it was painted:
18.75 ft² ÷ 2 = 9.375 ft²
Now it's left:
18.75 ft² - 9.375 ft² = 9.375 ft²

Before the fourth break, it was painted:
9.375 ft² ÷ 2 = 4.6875 ft²
Now it's left:
9.375 ft² - 4.6875 ft² = 4.6875 ft²

Now, we will sum what he painted for now:
75 ft² + 37.5 ft² + 18.75 ft² + 9.375 ft² 4.6875 ft² = 141.3125 ft² ≈ 141.35 ft²

When the painter takes his fifth break, there will be 141.35 ft² of the wall painted.

6 0

1 year ago

Read 2 more answers

The length of life of an instrument produced by a machine has a normal ditribution with a mean of 12 months and standard deviati

Zielflug [23.3K]

Answer:

a) $P(X$

And we can find this probability using the normal standard distirbution or excel and we got:

$P(z$

b) $P(7$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

a) less than 7 months.

Let X the random variable that represent the length of life of an instrument of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,2)$

Where $\mu=12$ and $\sigma=2$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard distirbution or excel and we got:

$P(z$

b) between 7 and 12 months.

$P(7$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

8 0

1 year ago

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev

Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 30.05 inches

$\sigma$ = standard deviation = 0.2 inches

n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches) = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } }$ ) = P(Z < 3) = 0.99865

P(X bar <= 30.25) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } }$ ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches) = 0.99865 - 0.97725

= 0.0214

7 0

2 years ago

Suppose you are told that a distribution is said to be approximately normal because outliers have skewed the data. List the poss

andrew-mc [135]

Answer:

Step-by-step explanation:

Outline are values which are entirely different from those remaining values in a data set. These extreme values can skew an approximately normal distribution by skewing the distribution in the direction of the outliers and this makes it difficult for the data set to be analyzed.

Its effect is such that the mean becomes extremely sensitive to extreme outliers making it possible that the mean is this not a representative of the population and this theoretically affects the standard deviation.

8 0

1 year ago