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3 years ago

8

Explain why estimation is an important skill to know when adding and subtracting decimals.

2 answers:

salantis [7]3 years ago

7 0

I think personally it is way easier for me to do it that way. Let's say we have 6.687-4.798, then think how long it takes to solve this problem versus just solving 6.700-4.800. Does that make sense?

Alla [95]3 years ago

3 0

Estimation is a important skill to know when adding and subtracting decimals because if you estimate you have an idea on what your answer should be close to.

For example 4.68-2.38 could be 5-2 which is 3. 4.68-2.38 equals 2.3. 2.3 is close to 3.

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You are a sportswriter for a newspaper. Today you are writing an article about a professional basketball player. During the play

Paha777 [63]

For this case we can make the following rule of three:
1507 ----> 100%
1213 ----> x
Clearing the value of x we have:
x = (1213/1507) * (100)
x = 80.4910418
Rounding off we have:
x = 80.5%
Answer:
this player has made 80.5% of free throws

4 0

3 years ago

Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of t

bogdanovich [222]

Answer:

a) $\sigma_{\bar x} = 1.414$

b) $\sigma_{\bar x} = 1.414$

c) $\sigma_{\bar x} = 1.414$

d) $\sigma _{\bar x} = 1.343$

Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation =10

Size of the sample n = 50

a) The population size is infinite;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

c) When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }$

$\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }$

$\sigma _{\bar x} = 1.343$

7 0

3 years ago

Question 3 of 35

natka813 [3]

The answer is d I’m pretty sure

3 0

3 years ago

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Quadrilateral ABCD is dilated by a scale factor of 1/3 centered around (1,2). Which statement is true about the dilation

morpeh [17]

Answer:

I cannot help answer this question if you don't add the statements...

Step-by-step explanation:

5 0

3 years ago

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Common denominator of 2/5 and 1/10

olga2289 [7]

The common denominator of 2/5 and 1/10 is 10

3 0

3 years ago

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