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DiKsa [7]
3 years ago
7

A marble factory makes single marbles, then puts them in boxes of 10, cartons of 100, and crates of 1,000. How can the marble fa

ctory package 21,650 marbles?
Mathematics
1 answer:
Tju [1.3M]3 years ago
3 0

Given, a marble factory makes single marbles and then puts them in boxes of 10, cartons of 100 and crates of 1000.

There are 21,650 marbles. We have to find how many boxes, cartons and crates are there.

We have to write 21,650 in the form of 1000, 100 and 10.

We can write 21,650 in the form of 1000 as, (21)(1000) = 21000

Now there (21650-21000) = 650 left.

We can write 650 in the form of 100 as, (6)(100) = 600

So there, (650-600) = 50 left.

Now we can write 50 in the form of 10 as, (5)(10) = 50.

So we can write 21,650 as,

21650 = (21)(1000)+(6)(100) + (5)(10)

So there are 21 crates, 6 cartons and 5 boxes in the marble factory package.

We have got the required answer here.

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Point C ∈ AB , AB = 30m. Point C is 1.5 times farther from point A than from point B. Find AC and CB.
Shtirlitz [24]

let's say that C is "x" units farther from B, that means that CB = x, and therefore AC = 1.5x.


\bf \underset{\leftarrow ~~30~~\to}{\boxed{A}\stackrel{1.5x}{\rule[0.35em]{18em}{0.25pt}}C\stackrel{x}{\rule[0.35em]{10em}{0.25pt}}\boxed{B}}
\\\\\\
AB=AC+CB\implies \stackrel{AB}{30}=\stackrel{AC}{1.5x}+\stackrel{CB}{x}\implies 30=2.5x
\\\\\\
\cfrac{30}{2.5}=x\implies 12=x
\\\\[-0.35em]
~\dotfill\\\\
AC=1.5(12)\implies AC=18~\hspace{8em} CB=x\implies CB=12

8 0
3 years ago
A mechanic had a bolt with a diameter of 2/9 inch. Will the bolt fit into a hole with a diameter of 0.2 inch.
Naddik [55]

Answer:

No, the bolt won't fit into the hole

Step-by-step explanation:

We need to convert the diameter of the bolt into decimal by dividing:

2/9 = 0.222...

The hole has diameter of 0.2

<em>Hence, 0.222 is larger than 0.2, so the bolt won't fit.</em>

3 0
3 years ago
Move two choices to the blanks to correctly complete the sentences.
Pavel [41]

Answer:

(A) There should have been 5 outcomes of HT

(B) The experimental probability is greater than the theoretical probability of HT.

Step-by-step explanation:

Given

S = \{HH,HT,TH,TT\} -- Sample Space

n(S) = 4 --- Sample Size

Solving (a); theoretical outcome of HT in 20 tosses

First, calculate the theoretical probability of HT

P(HT) = \frac{n(HT)}{n(S)}

P(HT) = \frac{1}{4}

Multiply this by the number of tosses

P(HT) * n= \frac{1}{4} * 20

P(HT) * n= 5

<em></em>

Solving (b); experimental probability of HT

Here, we make use of the table

P(HT) = \frac{n(HT)}{n(S)}

P(HT) = \frac{6}{20}

P(HT) = 0.30 ---- Experimental Probability

In (a), the theoretical probability is:

P(HT) = \frac{1}{4}

P(HT) = 0.25 ---- Experimental Probability

By comparison;

0.30 > 0.25

6 0
3 years ago
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Yes. Since 12 3/4 cups can be rounded up to 13 and 6 5/6 can be rounded up to seven, and 13+7 equals 20, so you would be able to put the additional amount if cups into the cooler.
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