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kipiarov [429]
3 years ago
5

George and Carmen went on a bicycle trip. they took a bus to their starting point, and then biked the rest. They traveled 3oo ki

lometers in total, and they biked 70 kilometers more than they were bussed.
Mathematics
1 answer:
jeka943 years ago
8 0

Answer:

George and Carmen traveled 185 kilometers by bike and 115 kilometers by bus

Step-by-step explanation:

<u><em>The question is</em></u>

Find how many kilometers they traveled by bike

Let

x ---->  kilometers traveled by bike

y ---->  kilometers traveled by bus

we know that

They traveled 300 kilometers in total

so

x+y=300 ----> equation A

They biked 70 kilometers more than they were bus

so

x=y+70 ----> equation B

substitute equation B in equation A

(y+70)+y=300

solve for y

2y+70=300\\2y=300-70\\2y=230\\y=115

<em>Find the value of x</em>

x=115+70=185

therefore

George and Carmen traveled 185 kilometers by bike and 115 kilometers by bus

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Answer:

Step-by-step explanation:

F(x) = 7 - 5x + 7x²

f(-9) = 7 - 5*(-9) + 7*(-9)²

       = 7 + 45 + 7*81

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7 0
3 years ago
Read 2 more answers
in a survey of 7000 women 4431 say they change their nail polish once a week.Construct a 95% confidence interval for the populat
andreyandreev [35.5K]

Answer:

The 95% confidence interval would be given (0.622;0.644).

We are confident (95%) that the true proportion of people that said that they change their nail polish once a week  is between 0.622 and 0.644

Step-by-step explanation:

Data given and notation  

n=7000 represent the random sample taken    

X=4431 represent the people that said that they change their nail polish once a week

\hat p=\frac{4431}{7000}=0.633 estimated proportion of people that said that they change their nail polish once a week

\alpha=0.05 represent the significance level

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Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.633 - 1.96 \sqrt{\frac{0.633(1-0.633)}{7000}}=0.622

0.633 + 1.96 \sqrt{\frac{0.633(1-0.633)}{7000}}=0.644

And the 95% confidence interval would be given (0.622;0.644).

We are confident (95%) that the true proportion of people that said that they change their nail polish once a week  is between 0.622 and 0.644

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