What layer in the atmosphere has the greatest depth and how deep is it?

What is the length of leg sof the triangle below? 90°

miskamm [114]

Not sure I’m sorry, have you tried google

5 0

3 years ago

Y=-x-3

nadya68 [22]

Answer: Heyaa! :)

1. y=-x-3 = -1

Slope: −1

y-intercept: (0,−3)

2. y=4x-5 = 4

Slope: 4

y-intercept: (0,−5)

3. 2x-y=-2 = 2

Slope: 2

y-intercept: (0,2)

4. x+y=4 = 1

Slope: −1

y-intercept: (0,4)

5. 3x+4y=-12 = - 3/4

Slope: −3/4

y-intercept:

(0,−3)

Hopefully this helps you!

- Matthew

4 0

1 year ago

A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation

Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

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

Normal probability distribution

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.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03$

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

$Z = \frac{3.1 - 2.2}{0.03}$

$Z = 30$

$Z = 30$ has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0

3 years ago

How does this polynomial identity work on numerical relationships? (y + x) (ax + b)

Serggg [28]

Let us take 'a' in the place of 'y' so the equation becomes

(y+x) (ax+b)

Step-by-step explanation:

Step 1:

(a + x) (ax + b)

Step 2: Proof

Checking polynomial identity.

(ax+b )(x+a) = FOIL

(ax+b)(x+a)

ax^2+a^2x is the First Term in the FOIL

ax^2 + a^2x + bx + ab

(ax+b)(x+a)+bx+ab is the Second Term in the FOIL

Add both expressions together from First and Second Term

= ax^2 + a^2x + bx + ab

Step 3: Proof

(ax+b)(x+a) = ax^2 + a^2x + bx + ab

Identity is Found .

Trying with numbers now

(ax+b)(x+a) = ax^2 + a^2x + bx + ab

((2*5)+8)(5+2) =(2*5^2)+(2^2*5)+(8*5)+(2*8)

((10)+8)(7) =(2*25)+(4*5)+(40)+(16)

(18)(7) =(50)+(20)+(56)

126 =126

3 0

3 years ago

What is the value of r6 + y2 when r =1 and y=11

Sergio039 [100]

1•6 + 11•2
6+22
28 would be the answer

7 0

3 years ago

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