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Len [333]
3 years ago
10

4. Given the line 7x + 6y = 5, do the following.

Mathematics
1 answer:
Annette [7]3 years ago
7 0

Answer:

a- Slope-intercept form (y =  \frac{-7 x}{6\\} + \frac{5}{6})  ::=> 1

b- Slope of Line =  m = \frac{-7}{6}    : From Equation 1

Step-by-step explanation:

a- We have following equation:

7x + 6y = 5

Here if we re-arrange for y:

6y = 5 -7x  :: Compare it with y = mx + b

y =  \frac{-7 x}{6\\} + \frac{5}{6}

b-Compare  y = mx + b and y =  \frac{-7 x}{6\\} + \frac{5}{6}

m= -7/6 and Slope of lines parallel to the given line is shown in the equation

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According to the graph it never crosses the x axis meaning that it has 0 roots beecasue roots is another word for x-values.

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a) The cost for 10 mile taxi ride is $18

b) The cost for m mile taxi ride is 1.5 m + 3

Step-by-step explanation:

Step 1 :

The fixed charges for the pick up = $3

Charges per mile = $1.50

Let s denote the total miles driven and t be the total cost for the trip

This can be represented by the equation

t = 3 + 1.5s

Step 2:

Distance traveled by Jonathan in his trip = 10 miles

So cost for riding 10 miles is

t = 3 + 1.5(10) = 3 + 15 = $18

The cost for 10 mile taxi ride is $18

Step 3 :

If the distance traveled is m miles, then substituting s = m in the above equation we get the cost as 1.5 m + 3

Step 4 :

Answer :

a) The cost for 10 mile taxi ride is $18

b) The cost for m mile taxi ride is 1.5 m + 3

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3 years ago
According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bother
k0ka [10]

Answer:

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

p_v =P(z>7.124)=5.24x10^{-13}  

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 proportion of interest is significantly higher than 0.5.  

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

\hat p=\frac{331}{502}=0.659 estimated proportion of people who would not be bothered if the NAtional security agency

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than the majority of 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, 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.

Calculate the statistic  

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

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

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=0.01. 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>7.124)=5.24x10^{-13}  

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 proportion of interest is significantly higher than 0.5.  

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