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erica [24]
3 years ago
6

What is the value of 2 in 5,670,249,114? And describe 2 ways to find value?

Mathematics
1 answer:
kirill115 [55]3 years ago
4 0
There are two ways in finding the place value of 2 in 5,670,249,114
 First by breaking down the numbers:
 5,670,249,114 = 5,000,000,000 + 600,000,000 + 70,000,000 + 200,000 + 40, 000 + 9,000 + 100 + 10 + 4  which means the Value of 2 is 200, 000<span>2. Second, by using the ones words:
5,670,249,114 = 5 billion + 6 hundred million + 70 ten million + 2 hundred thousand + 4 ten thousand + 9 thousands + 1 hundreds + 1 tens + 4 ones  which means the Value of 2 is 2 hundred thousand

</span>



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For which equation is m = 12 the solution? a. 4m=40 b. m+20=42. c. 4m=48. d. m-4=9
Gelneren [198K]
To solve this question, we just need to insert 12 into the m position of each question and see if the equation holds true.

a. 4m = 40

4(12) = 40

48 = 40

48 obviously does not equal 40, so it is not choice A.

b. m + 20 = 42

12 + 20 = 42

32 = 42

Again, 32 isn't the same as 42, so not choice B either.

c. 4m = 48

4(12) = 48

48 = 48

It looks like this one is true, but let's solve D also just to make sure.

d. m - 4 = 9

12 - 4 = 9

8 = 9

This is false, since 8 does not equal 9.

Therefore, choice C (4m = 48) is the correct answer.
Hope that helped! =)
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List 3 values that would make this inequality true 8+y&gt;11
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A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
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