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2 years ago

USE A MODEL Olivia and her brother William had a bicycle race. Olivia rode at a speed

Mathematics

1 answer:

Anni [7]2 years ago

7 0

Answer:

Let the total distance Aileen ran be 'x'

Let the distance ran by Andrew be 'y'

Refer to graph attached for graphical representation of distance travelled by both.

Applying the formula,

distance=speed*time

x=20*t Equation 1

y= 15*t equation 2

Since it can be easily understood by the graph attached, that

x=150+y equation 3

Now putting values of x and y in equation 3

20*t =150 +15*t.

5t= 150

t=30 seconds.

putting the values of t in both the equation 1 and 2

x=20t

=20*30

x = 600 feet

So the distance ran by Aileen is 600 feet, whereas Andrew ran 450 feet.

Step-by-step explanation:

Do not mind the names just change them. I had the same question before with different names so that is why. Hope this helps:)

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A mother earned $8750.00 from royalties on her cookbook. She set aside 20% of this for a down payment on a new home. The balan

Masteriza [31]

The problem statement is not clear regarding which money is invested. We will assume it is the 80% of $8750 that is set aside for future education. That amount is

... 0.80 × $8750 = $7000

If x represents the amount invested at 7%, then the total interest earned is

... (7000 -x)×4% + (x)×7% = 420

... 280 +0.03x = 420

... x = 140/0.03 ≈ 4666.67

$4666.67 was invested at 7%.

$2333.33 was invested at 4%.

5 0

3 years ago

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

$Z = \frac{X - \mu}{s}$

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

$Z = \frac{X - \mu}{s}$

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

$Z = \frac{X - \mu}{s}$

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

$Z = \frac{X - \mu}{s}$

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

2 years ago

How do I know what the circle and triangle equal to?

Brrunno [24]

Answer:

circle= 5, triangle= 3

Step-by-step explanation:

Please see attached picture for full solution.

8 0

2 years ago

National Income (NI) is the sum of the incomes that all individuals in an economy earn in the forms of wages,

Dennis_Churaev [7]

Answer:

\simeq 14.94 billion dollars

Step-by-step explanation:

During the period 1994 - 2004, the 'National Income' ,(NI) of Australia grew about 5.2% per year (measured in 2003 U. S, dollars). In 1994 , the NI of Australia was $ 4 billion.

Now,

(2020 - 1994) = 26

Assuming this rate of growth continues, the NI of Australia in the year 2020 (in billion dollars) will be,

$4 \times[\frac{(100 + 5.2)}{100}}]^{26}$

= $4 \times[\frac{105.2}{100}]^{26}$

=\simeq 14.94 billion dollars (answer)

8 0

3 years ago

What is the length of one leg of the triangle? 64 cm 64 StartRoot 2 EndRoot cm 128 cm 128 StartRoot 2 EndRoot cm.

zmey [24]

Answer:

Use the Pythagorean theorem for right triangles to solve this

a^2 + b^2 = c^2 where 'a' and 'b' are the legs of the triangle and 'c' is the hypotenuse

so:

8^2 + b^2 = 17^2 solve for the second leg, b

64 + b^2 = 289

b^2 = 289 - 64= 225

b = 15

7 0

2 years ago