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KatRina [158]
3 years ago
11

Read the following text.

Mathematics
1 answer:
bixtya [17]3 years ago
7 0

Answer:

The answer is 1:They describe as being a strong and heroic type of character

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The current value of a new car is $18,000. The car will depreciate 15% per year over the next 5 years. Which exponential equatio
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Answer:

f(x) = 18,000 • 0.85^x

Step-by-step explanation:

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Verbal contracts are generally discouraged because
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b

Step-by-step explanation:

sorry if wrong

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Use mental math and a pattern to find the product of 70*300
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70*300

=70(100+100+100)

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Answer the 3 screenshotted questions below. 30 points!
Triss [41]

Answer:

13.  36a^2b^4π  

14. 16x^3y^4π

15. -\frac{c^3}{4a^7b^3}

Step-by-step explanation:

13.

A = π * r^2

r = 6ab^2

A = π*  (6ab^2)^2

A = π * (6ab^2)(6ab^2)

A = π * 36a^2b^4

A = 36a^2b^4π  

14.

V = π * r^2 * h

r = 2x

h = 4xy^4

V = π * (2x)^2 * 4xy^4

V = π * 4x^2 * 4xy^4

V = π * 16x^3y^4

V = 16x^3y^4π

15. \frac{3a^{-4}b^2}{-12a^3b^5c^{-3}} = -\frac{c^3}{4a^7b^3}

\frac{3a^{-4}b^2}{-12a^3b^5c^{-3}}

\frac{a^{-4}}{-4a^3b^5c^{-3}}

\frac{b^2}{-4^{3+4}b^5c^{-3}}

\frac{b^2}{-4^7b^5c^{-3}}

\frac{1}{-4a^7b^{5-2}c^{-3}}

\frac{1}{-4a^7b^3c^{-3}}

\frac{c^3}{-4a^7b^3}

-\frac{c^3}{4a^7b^3}

Hope this helps!

3 0
3 years ago
Assume that the one-way commute time of an UoU student from his house to school is a normally distributed random variable which
Marizza181 [45]

Answer:

n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16

So the answer for this case would be n=16 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=10 represent the sample standard deviation

n represent the sample size  

ME=5 represent the margin of error

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (2)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (3)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.96, replacing into formula (3) we got:

n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16

So the answer for this case would be n=16 rounded up to the nearest integer

3 0
3 years ago
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