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VikaD [51]
1 year ago
10

Becca and Kimberly are running at a constant speed in a marathon. Becca runs at 4.5 miles per hour. Kimberly’s progress is shown

in the table.
Part A: Who runs faster?

Part B: If Becca and Kimberly are 5 hours into a marathon, how far has each run?

Mathematics
1 answer:
Georgia [21]1 year ago
5 0

Answer:

Part A: Kimberly

Part B:

Becca = 22.5 miles, Kimberly = 25 miles

Step-by-step explanation:

Part A:

Given that Becca runs at a constant speed of 4.5 miles per hour, use the table to find the constant speed that Kimberly runs, and compare who runs faster.

Let x = time

y = distance

k = speed = y/x

Equation for each person can be written as: y = kx

Therefore:

✔️Equation for Becca if k = 4.5

y = 4.5x

✔️Find k (speed) of Kimberly using (2, 10):

Speed (k) = y/x

k = 10/2

k = 5 miles per hour

Equation for Kimberly would be:

y = 5x

Comparing their speed, Kimberly runs faster because she covers more miles per hour than Becca does.

Part B:

For Becca, substitute x = 5 into Becca's equation, y = 4.5x

Thus:

y = 4.5*5 = 22.5 miles

For Kimberly, substitute x = 5 into Kimberly's equation, y = 5x

Thus:

y = 5*5 = 25 miles

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The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci
Olenka [21]

Answer:

The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}

For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that n = 300 and \pi = \frac{10}{300} = 0.033

Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532

The correct answer is

(0.0128, 0.0532)

4 0
2 years ago
Nikki drew a rectangle with a perimeter of 18 units on a coordinate grid. Two of the vertices were (4, –3) and (–1, –3). What co
Talja [164]

Answer:

There are two possible solutions for the other two vertices of the rectangle:

(i) (4, 1), (-1, 1), (ii) (4, -7), (-1, -7)

Step-by-step explanation:

Geometrically speaking, the perimeter of a rectangle (p) is:

p = 2\cdot b + 2\cdot h (1)

Where:

b - Base of the rectangle.

h - Height of the rectangle.

Let suppose that the base of the rectangle is the line segment between (4, -3) and (-1, -3). The length of the base is calculated by Pythagorean Theorem:

b = \sqrt{[(-1)-4]^{2}+[(-3)-(-3)]^{2}}

b = 5

If we know that p = 18 and b = 5, then the height of the rectangle is:

2\cdot h = p-2\cdot b

h = \frac{p-2\cdot b}{2}

h = \frac{p}{2}-b

h = 4

There are two possible solutions for the other two vertices of the rectangle:

(i) (4, 1), (-1, 1), (ii) (4, -7), (-1, -7)

8 0
2 years ago
Bryan’s golf coach suggested he take some golf lessons. The Pro at Windy Fairways charges $20.00 per month plus $10.00 per lesso
Aneli [31]

Bryan has to take 8 classes for both Pro's to be the same price.

Step-by-step explanation:

Given,

Monthly charges of Pro at Windy = $20.00

Charges per lesson = $10.00

Let,

x be the number of lessons

W(x) = 10x +20

Monthly charges of Sunny Sands = $100.00

They offer unlimited classes.

S(x) = 100

For the price to be same;

W(x) = S(x)

10x+20=100\\10x=100-20\\10x=80

Dividing both sides by 10

\frac{10x}{10}=\frac{80}{10}\\x=8

Bryan has to take 8 classes for both Pro's to be the same price.

Keywords: function, division

Learn more about division at:

  • brainly.com/question/12012120
  • brainly.com/question/12041380

#LearnwithBrainly

5 0
2 years ago
Manuela solved the equation 3−2|0.5x+1.5|=2 for one solution. Her work is shown below. 3−2|0.5x+1.5|=2 −2|0.5x+1.5|=−1 |0.5x+1.5
lions [1.4K]

Answer:

Step-by-step explanation:

We'll just work on solving both so you can see what's involved in solving an absolute value equation. Because an absolute value is a distance, we can have that distance being both to the right on the number line of the number in question or to the left. For example, from 2 on the number line, the numbers that are 5 units away are 7 and -3. Using that logic, we will simplify the equation down so we can set up the 2 basic equations needed to solve for x.

If  3-2|.5x+1.5|=2 then

-2|.5x+1.5|=-1  What you need to remember here is that you cannot distribute into a set of absolute values like you would a set of parenthesis. The -2 needs to be divided away:

|.5x+1.5|=.5

Now we can set up the 2 main equations for this which are

.5x + 1.5 = .5  and .5x + 1.5 = -.5

Knowing that an absolute value will never equal a negative number (because absolute values are distances and distances will NEVER be negative), once we remove the absolute value signs we can in fact state that the expression on the left can be equal to a negative number on the right, like in the second equation above.

Solving the first one:

.5x = -1 so

x = -2

Solving the second one:

.5x = -2 so

x = -4

7 0
2 years ago
Read 2 more answers
Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly s
kifflom [539]

Answer:

The expected number of tests, E(X) = 6.00

Step-by-step explanation:

Let us denote the number of tests required by X.

In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.

To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.

Case 1: P(X=1) = [P (not infected)]⁵

                       = (0.15 - 0.1)⁵

            P(X=1) = 3.125*10⁻⁷

Case 2: P(X=6) = 1- P(X=1)

                        = 1 - (1 - 0.1)⁵

               P(X=6) = (1 - 3.125*10⁻⁷) = 0.999999

               P(X=6) = 1.0

We can then use the previously determined values to compute the expected number of tests.

E(X) = ∑x.P(X=x)

      = (1).(3.125*10⁻⁷) + 6.(1.0)

 E(X)  =  E(X) = 6.00

Therefore, the expected number of tests, E(X) = 6.00

3 0
2 years ago
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