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2 years ago

Jose prefers to walk to work when the weather

Mathematics

1 answer:

stich3 [128]2 years ago

6 0

Answer:

independent: time (hours)

dependent: distance (miles)

Step-by-step explanation:

The independent variable would be time because it can be changed depending on his walking speed.

The dependent variable would be distance

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9. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of

netineya [11]

Answer:

9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 12, \sigma = 3$

What is the probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated?

This is 1 subtracted by the pvalue of Z when X = 16. So

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

$Z = \frac{16 - 12}{3}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

1 - 0.9082 = 0.0918

9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated

5 0

3 years ago

Write a rule for the nth term of the arithmetic sequence. -23,-30,-37,-44...

n200080 [17]

You can try khan academy

8 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago

3/4 -3/8 in simplest form?

solmaris [256]

3/4 and 3/8 have to have equal denominators, so multiply 3/4 by 2/2 because it's equal to one, so it doesn't change the value of the fraction, but it makes the denominator (the number on the bottom) bigger.
3/4 x 2/2 = 6/8
Now subtract.
6/8 - 3/8 = 3/8
Don't subtract the denominators.
3/8 can't be simplified.

7 0

3 years ago

Read 2 more answers

You have been asked to build a scale model of your school out of toothpicks. imagine your school is 30 feet tall. your scale is

sveticcg [70]

Since your scale is 1ft:1.26cm, a 30-ft tall school would need to have a $30*1.26=37.8$ cm model. Dividing this by how tall each toothpick is, you'll get:

$\frac{37.8}{6.3} =6$

ANSWER: The model would be 6 toothpicks tall.

To find out how many cotton swabs you'll need, we just divide 37.8 by how tall each swab is:

$\frac{37.8}{7.7} =5$

ANSWER: The model would be 5 cotton swabs tall.

4 0

3 years ago