How do I multiply whole fractions

A person suspects a standard number cube is not so standard. Another person rolls this cube 100 times, and it lands on a six 42

Ksju [112]

Answer:

The chance of an average six-sided number cube landing one one number is 1/6, or approximately 16.67%. When rolling a number cube 100 times, the average number of times it lands on any specific number should be fairly close to 16.67.

A 6 being rolled 42 times and 33 times means that it is very likely not to be a standard number cube. Although we cannot know for certain, the chance of this happening is very low.

3 0

3 years ago

At the local gas station gasoline costs $2.75 for per gallon and diesel costs $3.02 per gallon. Gabriel needs to purchase 15.2 g

Ainat [17]

Answer:

The total amount paid by Gabriel at the local gas station is $70.49.

Step-by-step explanation:

Cost of gasoline per gallon = $2.75

Cost of diesel per gallon = $3.02

Amount of gasoline Gabriel needs = 15.2 gallons

Amount of diesel Gabriel needs = 9.5 gallons

Compute the total amount paid by Gabriel at the local gas station as follows:

Total Cost = Cost of gasoline × Amount of gasoline

+ Cost of diesel × Amount of diesel

$=(2.75\times 15.2)+(3.02\times 9.5)\\\\=41.8+28.69\\\\=70.49$

Thus, the total amount paid by Gabriel at the local gas station is $70.49.

7 0

3 years ago

What is the slope of the line y= 3?

Kipish [7]

Answer:

Undefined

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

How can you use the measures of center and the range to compare two populations?

DochEvi [55]

The measure of center and range to be used to compare the population is the median and the inter quartile range.

Explanation:

You utilize the median and the interquartile range (IQR) to portray skewed distributions of information. To look at two populaces, utilize the mean and the MAD when the two populations are symmetric. Utilize the middle and the IQR when possibly one or the two circulations are slanted.

In statistics and probability theory, the median is the worth isolating the higher half from the lower half of an information test, a populace or a likelihood circulation. For an informational collection, it might be thought of as the "center" esteem.

8 0

3 years ago

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

1. A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin; 2. The radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball; 3. The volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

or

$\\ 1 \leq a < 10$ (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The La Plata river dolphin weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the scientific notation):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin.

<h3>Second case</h3>

Following the same reasoning, we can conclude that the radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball:

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$