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3 years ago

Express 3m24cm in kilometers,leaving your answer in standard form

Mathematics

2 answers:

erastova [34]3 years ago

7 0

Answer:

0.00324 kilometres

Step-by-step explanation:

Remember:

100 centimeters = 1 meter

1000 meters = 1 kilometer

So first, let's add together the meters and centimeters. To do that, we have to convert the meters to centimeters. That would become 300 centimeters, then you add the 24 centimeters to get 324 centimeters. Then, you convert that to kilometers. You can use a calculator for this, but I'll show you how to do this without a calculator. Since 100 centimeters is one meter, and 1000 meters equals to one kilometer, there would be 5 spaces from 324 to our answer, so you move the invisible decimal point 5 spaces to the left, and you get your answer, which is 0.00324 kilometres!

Jlenok [28]3 years ago

3 0

Answer:

0.00324 km is the answer

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pentagon [3]

The answer is 2√19 I think

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2 years ago

How do you do standard form, expanded form and words with decimals 2.07?

Karo-lina-s [1.5K]

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3 years ago

Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let X = number of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, p = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, n = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable X follows a Binomial distribution with parameter n = 55 and p = 0.20.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:

$E(X)=np$

$=55\times 0.20\\=11$

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$

$=0.1968$

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0

3 years ago

URGENT...

MatroZZZ [7]

The center of the dilation is the origin, and the scale factor is 2.

To find the coordinates of the points under this transformation, it is enough to multiply both coordinates of the points X, Y, and Z by 2.

Take a look at the dilation of point Z. To get to Z', both the x and y coordinate of Z must be multiplied by 2.

So, the coordinates are:

X'(8, 0); Y'(6, 4); Z'(4, -4).