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lesya692 [45]
3 years ago
13

Miguel rode his bicycle 4 miles less than 5 times the number of miles that Nathan rode. If Miguel rode his bicycle 6 miles, how

many miles did Nathan ride?
Mathematics
2 answers:
salantis [7]3 years ago
4 0
5n-4 = 6 gives the equation for 5 times #Nathan rode less 4 miles which solves to n = 2
Nathan rode 2 miles. (5*2)-4 = 6   10-4=6 so 2 miles is right
Zanzabum3 years ago
4 0

Answer:

Nathan rode his bike 2\ miles

Step-by-step explanation:

Let

x-----> number of miles that Miguel rode his bike

y-----> number of miles that Nathan rode his bike

we know that

x=5y-4 -----> equation A

x=6\ mi -----> equation B

equate equation A and equation B and solve for y

6=5y-4

5y=6+4

5y=10

y=2\ mi

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Simplify the expression: 2(2 + g) =
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Answer:

4 + 2g

General Formulas and Concepts:

<u>Pre-Algebra</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

2(2 + g)

<u>Step 2: Expand</u>

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Multiply:<br> x (x^2+2x+4) and -2 (x^2+2x+4)<br><br> Combine like terms.<br><br> Final Answer:
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Answer:

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The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from
Vinvika [58]

Answer:

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

df=n_1 +n_2 -2=13+19-2=30  

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

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

\bar X_{1}=810 represent the mean for sample 1  

\bar X_{2}=770 represent the mean for sample 2  

s_{1}=67 represent the sample standard deviation for 1  

s_{2}=56 represent the sample standard deviation for 2  

n_{1}=13 sample size for the group 2  

n_{2}=19 sample size for the group 2  

\alpha=0.05 Significance level provided

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=13+19-2=30  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

P value  

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

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

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