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3 years ago

12

How many skittles candies can fit in a mason jar that has the inner dimensions that are approximately a cylinder with a height o

f 9 cm and a base radius of 3 cm

1 answer:

ololo11 [35]3 years ago

8 0

Silly question but do you have any information on the skittles? surely they would give you measurements for them too?

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Indicate below whether the equation in the box is true or false 1/8 = to 2/4 true or false

Semenov [28]

Answer:

false

Step-by-step explanation:

1/8 = .125

2/4 = .5

therefore 1/8 does not = 2/4

8 0

3 years ago

Compare the volume and surface area of a sphere with a radius of 2 cm with that of a cube that has sides measuring 3.22 cm

Naddik [55]

Comparision of volume of sphere and cube=0.98820059
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3 years ago

Someone help please!

Lesechka [4]

X is equal to 82 degrees

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3 years ago

A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with

Ratling [72]

Answer:

$P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)$

And we can find this probability using the complement rule and with excel or the normal standard table:

$P(z>1.61)=1-P(z$

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".

Solution to the problem

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

$X \sim N(10.5,0.3)$

Where $\mu=10.5$ and $\sigma=0.3$

We are interested on this probability

$P(X>10.983)$

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:

$P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)$

And we can find this probability using the complement rule and with excel or the normal standard table:

$P(z>1.61)=1-P(z$

8 0

3 years ago

Is 5266 divisible by 2

Gemiola [76]

Answer:

yes it is, the answer is 2633

3 0

3 years ago

Read 2 more answers