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SCORPION-xisa [38]
3 years ago
6

Evaluate 5/3

Mathematics
2 answers:
Helga [31]3 years ago
7 0

Answer:

120

Step-by-step explanation:

ziro4ka [17]3 years ago
4 0
120 is good engough if you dont kn o
You might be interested in
A mirror frame in the shape of an oval is shown below. The ends of the frame form semicircles:
Kruka [31]
You need to add up three things:
- two sides of 60" each (60*2=120")
- the perimeter of the upper semicircle (3.14*27/2=42.412")
- the perimeter of the lower semicircle (3.14*27/2=42.412")

After addition:
120+42.412+42.412=204.823"


I hope you got the idea!








3 0
3 years ago
A newspaper poll asked respondents if they trusted "eco friendly" labels on cleaning products. Out of 1000 adults surveyed, 498
Paul [167]

Answer:

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of respondents that trust these labels is not significanlty less than 0.5 or 50%.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=498 represent the adults that trust these labels

\hat p=\frac{498}{1000}=0.498 estimated proportion of respondents that trust these labels

p_o=0.5 is the value that we want to test

\alpha represent the significance level

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 proportion of respondents that trust these labels is at least 50%:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

When we conduct a proportion test we need to use the z statisitc, 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{0.498 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-0.126  

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 is not provided but we can assume it as \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of respondents that trust these labels is not significanlty less than 0.5 or 50%.  

7 0
3 years ago
Solution.<br> 6) -162-5-X<br> 8 9 10 11<br> 12<br> 14 15 16 17 18
statuscvo [17]
X ≥ 11 is the answer
7 0
3 years ago
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
3 years ago
You earn 50,000 per year,and paid 10 percent in taxes this year. the government increased the tax rate to 20 percent foe the nex
bulgar [2K]

Answer:

D

Step-by-step explanation:

he paid 5000 in taxes last year multiply by 2 10000

D is the answer sorry i dont have a proper explanation

4 0
3 years ago
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