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Tresset [83]
2 years ago
10

Nathan flew 3,547 miles from Canada to California during the first part of his trip. He flew 2,567 miles from California to Hawa

ii during the second part of his trip. Which is the difference in the number of miles Nathan flew between the first and second parts of his trip?
Mathematics
1 answer:
AURORKA [14]2 years ago
4 0

Difference in the number of miles Nathan flew between the first and second parts of his trip is 980 miles

<em><u>Solution:</u></em>

Given that Nathan flew 3,547 miles from Canada to California during the first part of his trip

He flew 2,567 miles from California to Hawaii during the second part of his trip

Therefore,

first part of his trip = 3547 miles

second part of his trip = 2567 miles

Difference in the number of miles Nathan flew between the first and second parts of his trip is given as:

difference = first part of his trip - second part of his trip

difference = 3547 - 2567 = 980

Therefore, the difference in number is 980 miles

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Answer:

Correct choice is B

Step-by-step explanation:

All given options represent quadratic function. Let the equation of this quadratic function be

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Then

1. f(0)=2=a\cdot 0^2+b\cdot 0+c\Rightarrow c=2;

2. f(1)=7.5=a\cdot 1^2+b\cdot 1+c\Rightarrow 7.5=a+b+2;

3. f(2)=12=a\cdot 2^2+b\cdot 2+c\Rightarrow 12=4a+2b+2.

Solve the system of two equations:

\left\{\begin{array}{l}a+b+2=7.5\\4a+2b+2=12\end{array}\right.\Rightarrow\left\{\begin{array}{l}a=5.5-b\\4(5.5-b)+2b=10\end{array}\right.

Then

22-4b+2b=10,\\ \\-2b=-12,\\ \\b=6,\\ \\a=5.5-6=-0.5.

Thus, the equation of the function is

f(x)=-\dfrac{1}{2}x^2+6x+2.

Note that

f(3)=-\dfrac{1}{2}\cdot 3^2+6\cdot 3+2=-4.5+20=15.5;

f(4)=-\dfrac{1}{2}\cdot 4^2+6\cdot 4+2=18.

8 0
2 years ago
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An article in Medicine and Science in Sports and Exercise "Maximal Leg-Strength Training Improves Cycling Economy in Previously
Hatshy [7]

Answer:

The 99% confidence interval for the mean peak power after training is [299.4, 330.6]

299.4\leq\mu\leq 330.6

Step-by-step explanation:

We have to construct a 99% confidence interval for the mean.

A sample of n=7 males is taken. We know the sample mean = 315 watts and the sample standard deviation = 16 watts.

For a 99% confidence interval, the value of z is z=2.58.

We can calculate the confidence interval as:

M-z\sigma/\sqrt{n}\leq\mu\leq M+z\sigma/\sqrt{n}\\\\315-2.58*16/\sqrt{7}\leq\mu\leq 315+2.58*16/\sqrt{7}\\\\315-15.6\leq \mu\leq 315+15.6\\\\299.4\leq\mu\leq 330.6

The 99% confidence interval for the mean peak power after training is [299.4, 330.6]

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2 years ago
A local cable company claims that the proportion of people who have Internet access is less than 63%. To test this claim, a rand
MaRussiya [10]

Answer:

Step-by-step explanation:

For the null hypothesis,

H0 : p = 0.63

For the alternative hypothesis,

Ha : p < 0.63

This is a left tailed test

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q = probability of failure = 1 - p

q = 1 - 0.63 = 0.37

Considering the sample,

Sample proportion, P = x/n

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P = 478/800 = 0.6

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

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<u>Given</u>:

The given balanced scale is represented by the equation

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<u>Process to balance the scale:</u>

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Thus, the equation (3) is the same as the equation (1).

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If Cassie has 20 and wants to keep 11 then we would subtract 11 from 20.

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