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Lorico [155]
3 years ago
11

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
3 0

Answer:

dy/dx = 2/x

Step-by-step explanation:

y = In x^2

Let u = x^2

du/dx = 2x

y = In u

dy/du = 1/u

dy/dx = dy/du × du/dx = 1/u × 2x = 1/x^2 × 2x = 2/x

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1. A local pizzeria sold 65 pizzas yesterday. (a) The owner offers the employees a bonus if the number of pizzas sold increases
Ad libitum [116K]
I think the answer is

65 + 0.20(65) = 65 + 13 = 78..they would have to sell 78 to get the bonus

(65 - 60) / 65 = 5/65 = 0.0769 x 100 = 7.69 rounds to 7.7 % decrease compared to the 65 pizza's sold yesterday

Not completely sure though, but
Hope It Helps
6 0
3 years ago
If someone can help with this tysm, in class rn tryna get an A
Rom4ik [11]

Answer:

1. 4/9

2. 80,100,6.5

3.  10 and 11

4. 56cm

5. 3 an4

6. Real, Rational, Integer

7. All whole numbers are natural numbers

8. -1 1/2

9. √72

10. √9

Good luck :)

Step-by-step explanation:

6 0
3 years ago
According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

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

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

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

By the Central Limit Theorem:

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

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

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

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

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2 years ago
PLS HELP TIMED EXAM plsss
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Might be 134 because since angle 2 and 8 are lined up like that it is same side exterior angles which means they are supplementary from each other so they will equal 180. So I did 180-46 to get 134, I’m not sure if I’m correct
7 0
2 years ago
What is 4 to the 60th power
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What is 4th to the 60th piwer
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