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3 years ago

average age of 15 students of iub 11years if teacher is also included average age becomes 13 years how old is teachers

Mathematics

1 answer:

Yuki888 [10]3 years ago

3 0

Answer: the teacher is 43

Step-by-step explanation: if you take 11 and multiply it by 15 you get 165 if you take 208 and divide it by 16 you get 13.

so basically you subtract 208 from 165 to get 43

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Rewrite the function, differentiate, and then simplify y= 3/√x

Anit [1.1K]

Answer:

$\frac{dy}{dx}$ = - $\frac{3}{2x^{\frac{3}{2} } }$

Step-by-step explanation:

Differentiate using the power rule

$\frac{d}{dx}$ (a $x^{n}$ ) = na $x^{n-1}$

Given

y = $\frac{3}{\sqrt{x} }$ = 3 $x^{-\frac{1}{2} }$ , then

$\frac{dy}{dx}$ = - $\frac{3}{2}$ $x^{-\frac{3}{2} }$ = - $\frac{3}{2x^{\frac{3}{2} } }$

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3 years ago

According to a candy company, packages of a certain candy contain 1616% orange candies. Suppose we examine 100100 random candi

GaryK [48]

Answer:

a) By the Central Limit Theorem, 16%.

b) 0.0367 = 3.67%

c) We expect 16% orange candies, give or take 3.67%.

Step-by-step explanation:

The Central Limit Theorem estabilishes 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 $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

For samples of size n of a proportion p, the expected sample percentage is p and the standard error is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.16, n = 100$

a. What value should we expect for our sample percentage of orange candies?

By the Central Limit Theorem, 16%.

b. What is the standard error?

$s = \sqrt{\frac{0.16*0.84}{100}} = 0.0367$

0.0367 = 3.67%

c. Use your answers to fill in the blanks below. We expect ____% orange candies, give or take _____%.

We expect 16% orange candies, give or take 3.67%.

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3 years ago

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The <BAC angle isn't a perfect right angle because the image was a photo

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2 years ago

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Four times the sum of 14 and c is a squared

AlekseyPX

$4\times(14+c^2)=4\times14+4\times c^2=56+4c^2$

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3 years ago

The graph represents the normal distribution of recorded weights, in pounds, of cats at a veterinary clinic.

marshall27 [118]

Please consider the graph.

We have been given that graph represents the normal distribution of recorded weights, in pounds, of cats at a veterinary clinic. We are asked to choose the weights, which are within 2 standard deviations of the mean.

We can see from our graph that mean of the weights is 9.5 and standard deviation in 0.5.

The data point that would be below two standard deviation is: $9.5-2\times 0.5$ that is $9.5-1=8.5$ .

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Upon looking at our given choices, we can see that 8.9, 9.5 and 10.4 pounds lie within 2 standard deviation of the mean.

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3 years ago

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