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Kruka [31]
3 years ago
5

3=x+3-5x solving multi step I give Brainliest

Mathematics
2 answers:
Sergio039 [100]3 years ago
4 0

Answer:

x = 0

Step-by-step explanation:

3 = x + 3 - 5x

3 = -4x + 3

4x = 0

x = 0

brilliants [131]3 years ago
4 0

Answer:

x= 0

Step-by-step explanation:

combine the liked terms, then you subtract the 3 from both sides, then simplify

3=x+3-5x

3=-4x+3

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Simplify. Thank you...
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Answer:

I'm not entirely sure but I got:

\frac{ {4x}^{3} + 3 }{(2x - 1)( {x}^{2}  + x)( {2x}^{2} - 3x + 1) }

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In survey 17/25 of the people surveyed have a cat. what percent of the people surveyed have a cat?
mestny [16]

Answer:

The percentage of the people surveyed that have a cat is 68%

Step-by-step explanation:

we know that

To find the percentage of the people surveyed that have a cat, multiply the given fraction by 100

so

\frac{17}{25}*100=17*4=68\%

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3) You paid $5 for a bucket of popcorn and $1.25 per candy bar. How many candy bars
Alex73 [517]

4 dollars flat $4.00

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Help please and thanks! I usually give brainliest as well! :)
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2 years ago
A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe
d1i1m1o1n [39]

Answer:

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

df=n-1=15-1=14  

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation  

\bar X=5.63 represent the sample mean  

s=1.61 represent the standard deviation for the sample  

n=15 sample size  

\mu_o =15/tex] represent the value that we want to test  
[tex]\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 to be tested  

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:  

Null hypothesis:\mu \leq 5.63  

Alternative hypothesis:\mu > 5.63  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

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

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

Now we need to find the degrees of freedom for the t distribution given by:

df=n-1=15-1=14  

P value

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0
3 years ago
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