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masya89 [10]
3 years ago
11

Add 4 kg 250 g, 2 kg 90 g and 19kg 3 g.

Mathematics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

=25343 grams or 25 kg 343 g.

Step-by-step explanation:

1 kg = 1000 g

∴ 4 kg = 4000 grams

and the 4 kg 250 g would equal = 4250 grams.

∴ 2 kg = 1000 g

2 kg 90 g = 2000 grams + 90 g = 2090 grams

∴ 19 kg = 19000 grams

and 19 kg 3 grams = 19000 grams + 3 grams = 19003 grams

now we will add all of the values together to get the final answer.

4250 grams + 2090 grams + 19003 grams = 25343 grams or 25 kg 343 g.

Hope this helps!!

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The school that Stefan goes to is selling tickets to a choral performance. On the first day of ticket sales the school sold 3 se
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Answer: the price of a senior citizen's ticket is $8.

the price of a child's ticket is $14

Step-by-step explanation:

Let x represent the price of a senior citizen's ticket.

Let y represent the price of a child's ticket.

On the first day of ticket sales, the school sold 3 senior citizen tickets and 1 child ticket for a total of $38. It means that

3x + y = 38- - - - - - - - - - - -1

The school took in $52 on the second day by selling 3 senior citizen and 2 child tickets. It means that

3x + 2y = 52- - - - - - - - - - - -2

Subtracting equation 2 from equation 1, it becomes

- y = - 14

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Substituting y = 14 into equation 1, it becomes

3x + 14 = 38

3x = 38 - 14 = 24

x = 24/3

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3 years ago
Out of 431 applicants for a job, 142 have over 5 years of experience and 103 have over 5 years of experience and have a graduate
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Answer:

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Step-by-step explanation:

<u>Given</u>

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Probability = favorable outcomes/total outcomes

<u>Required probability </u>

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3 years ago
I geologist note is that the land alone at fault line moved 24.8 cm over the past 175 years on average how much did the land mor
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First, let's establish a ratio between these two values. We'll use that as a starting point. I personally find it easiest to work with ratios as fractions, so we'll set that up:

\frac{24.8\ cm}{175\ years}

To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:

\frac{24.8\ cm}{175\ years}\div  \frac{175}{175}  \approx  &#10;\frac{0.14\ cm}{1\ year}

We have our answer: approximately 0.14 cm or 1.4 mm per year
3 0
3 years ago
Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of
VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are 26^{5} = 11,881,336 of these words total

(b) How many of these words contain no repeated letters?

The first letter can be any of them, so L1 = 26.

At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

So we have

26-25-24-23-22

In total there are

26*25*24*23*22 = 7,893,600

of these words with no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

T = T_{1} + T_{2} + T_{3}

T_{1} is the number of words that start with an a and do not end with z. So we have

1 - 26 - 26 - 26 - 25

The first letter can only be a, and the last one cannot be z. So:

T_{1} = 26^{3}*25 = 439,400

T_{2} is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

T_{2} = 26^{3}*25 = 439,400

T_{3} is the number of words that both start with a and end with z. So:

1 - 26 - 26 - 26 - 1

The first letter can only be a, and the last can only be z. The other three letters could be anything. So:

T_{3} = 26^{3} = 17,576

T = T_{1} + T_{2} + T_{3} = 2*439,400 + 17,576 = 896,376

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