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shepuryov [24]
3 years ago
11

How many times doez 100,000,000 go into 1,000,000,000​

Mathematics
2 answers:
Neko [114]3 years ago
5 0

Answer:

100,000,000 goes 10 times into 1,000,000,000

igor_vitrenko [27]3 years ago
3 0
The answer to your question is ten
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Write an expression in expanded form and an expression in factored form for the diagram.
olga2289 [7]

Step-by-step explanation:

You put them together to form a bigger number then break it down. sorry if it doesn't make sense.

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What is the simplyfied fraction of 18/54
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Let f(x)=9x^3-12x^2-31 and let g(x)=3x-1. Find f(x)/g(x)
kumpel [21]

Answer: f(x)/g(x) = 3x^2 + 6x - 2 - 12/3x+1

Step-by-step explanation:

The first function is f(x) = 9x^3+21x^2-14

The second function is g(x) = 3x+1

     f(x)/g(x = 9x^3+21x^2-14/3x+1

We perform the long division as shown in the attachment to obtain the quotient as: Q(x) = 3x^2+6x - 2 and remainder R = -12

Therefore f(x)/g(x) = 3x^2 + 6x - 2 - 12/3x-1

   Where x does not = -1/3

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According to Inc, 79% of job seekers used social media in their job search in 2018. Many believe this number is inflated by the
Fynjy0 [20]

Answer:

Null hypothesis:p\leq 0.79  

Alternative hypothesis:p > 0.79  

z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786  

p_v =P(z>2.786)=0.00267  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.  

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=314 represent the respondents that use social media in their job search.

\hat p=\frac{314}{370}=0.849 estimated proportion of respondents that use social media in their job search

p_o=0.79 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is hgiher than 0.79.:  

Null hypothesis:p\leq 0.79  

Alternative hypothesis:p > 0.79  

When we conduct a proportion test we need to use the z statisticc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>2.786)=0.00267  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.  

7 0
3 years ago
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