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zvonat [6]
3 years ago
6

When asked to find a number one-tenth as large as another number, what operation would you use? What about when asked to find a

number 10 times as large?
Mathematics
1 answer:
kherson [118]3 years ago
6 0

Apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.

<u>Solution:</u>

Need to determine what operation is required to get one-tenth of a number and 10 times of a number

To get one tenth of a number, divide the number by 10.

For example to get one – tenth of 100, divide it by 10, we get 10 as a result.

\frac{1}{10} \times 100 = 10

To get ten times of a number, multiply the number by 10

For example 10 times of 10 = 10 x 10 = 100

Hence apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.

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Do you have choices?

The most simplified expression that is equal is
-27x-28
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Im very confuzed by this plz help me!
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Answer: 2.339

Step-by-step explanation:

Make sure that you align your decimal points. The 1 shouldn't line up with the 7.

Instead, write your equation like:

2.37

- 0.031

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I hope this helps! :)

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3 years ago
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What is the perimeter of a rectangle with two sides of 6inch and two sides of 8inch
yawa3891 [41]
P=28in because there are two sides that equal 6in and two other sides that equal 8. You multiply 6 and 8 by 2 which is 6*2=12 and 8*2=16. And then you add 12 and 16
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3 years ago
a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou
Fed [463]

Answer:

\displaystyle P(A)=\frac{6}{11}

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

\Omega=\{WW,WN,NW,NN\}

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

A=\{WN,NW\}

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

\displaystyle \frac{5}{11}

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

\displaystyle \frac{6}{10}

Thus the option WN has the probability

\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}

Now for the second option NW. The initial probability to pick a non-white counter is

\displaystyle \frac{6}{11}

The probability to pick a white counter is

\displaystyle \frac{5}{10}

Thus the option NW has the probability

\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}

The total probability of event A is the sum of both

\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}

\boxed{\displaystyle P(A)=\frac{6}{11}}

7 0
3 years ago
4400 dollars is place in an account with annual interest rate of 8.25% How much will be in the account after 22years to the near
tatiyna

Answer: A = 4400(1 + 0.0825 x 22)

Step-by-step explanation:

A = accumulated amount

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r = rate - 8.25% = 0.0825

t = years = 22

= 4400(1 + 0.0825 x 22)

6 0
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