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Marina86 [1]
2 years ago
12

How do you convert a terminating decimal into a fraction?​

Mathematics
1 answer:
AysviL [449]2 years ago
4 0
Write down the decimal divided by 1, like this: decimal. Multiply both top and bottom by 10 for every number after the decimal point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.). Lastly Simplify (or reduce) the fraction.
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Consider the positively charged particles seen here. The magnitude of their combined charges is represented by the vector arrow.
gavmur [86]

B)decreases by one; arrow length decreases by 1/3

3 0
3 years ago
Is 3/4 a rational number, whole number, integer?
Elden [556K]
I think its a rational number
4 0
2 years ago
Read 2 more answers
Glenn needs to cut pieces of ribbon that are each 1 meter long to make ribbon key chains. If he has 6 pieces of ribbon that are
Jet001 [13]

Answer:

2/3

Step-by-step explanation:

Glenn needs to cut pieces of ribbon that are each 1 meter long to make ribbon key chains. If he has 6 pieces of ribbon that are each 1 dekameter long, how many 1−meter pieces of ribbon can he cut?

1 decametre = 10 metres

Hence,

10 meters = 6 pieces of ribbon

1 meter = x pieces

Cross Multiply

10x = 6

x = 6/10

x = 2/3 (1 meter pieces of ribbon)

6 0
2 years ago
A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select
Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; \mu = 57,800  and  \sigma = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( \frac{X-\mu}{\sigma} <= \frac{58000-57800}{750} ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( \frac{X-\mu}{\sigma} < \frac{57000-57800}{750} ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = \frac{\sigma}{\sqrt{n} } = \frac{750}{\sqrt{100} } = 75

The z probability is given by;

           Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{58000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{57000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

7 0
3 years ago
Katerina is a botanist studying the production of pears by two types of pear trees. she noticed that Type A trees produced 20 pe
Ronch [10]

Answer:

120 pears

Step-by-step explanation:

Let the  number of pears produced by Type B be x

Number of pears  produced by type A=144

  According to question

Type A produce 20% more pears than B

x+ x ×20%= 144

x+x×\frac{20}{100}= 144

\frac{100x+20x}{100}= 144

\frac{120x}{100}=144

x= 144×\frac{100}{120}

x=120 pears

Hence, Number of pears produced by B= 120 pears

Hence, the correct answer is 120 pears

5 0
2 years ago
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