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4 years ago

Helppp I will give brainiest. I will have three more questions after.

Mathematics

1 answer:

lys-0071 [83]4 years ago

4 0

Answer:

The answer is 71.4

Step-by-step explanation:

357 / 5 = 71.4

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I need answer for this question

just olya [345]

The probability that the aircraft is overloaded is 97.98%, which means the pilot should take the action.

In a Normal distribution with mean ц and standard deviation σ, the z-score of a measure x is given by:

Z = X-ц / σ

· It measures how many standard deviations the measure is from the mean.

· After finding 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, the sampling distribution of sample means of the size n has standard deviation σ

σ = σ / $\sqrt{n}$

σ is standard deviation

n is the sample size.

Given that the mean and the standard deviation of the population is 176.1 lb and 35.4 respectively.

⇒ ц = 176.1 and σ = 35.4

For a sample of 43 passengers, we have

n = 43

σ = $\frac{35.4}{\sqrt{43}}$

σ = 5.398

Z = X-ц / σ

Z = $\frac{165-176.1}{5.398}$

Z = -2.05 has p- value of 0.9798

The probability that the aircraft is loaded is

1 - p-value of Z

1 - 0.0202 = 0.9798

The probability that the aircraft is overloaded is 97.98%

Know more about Normal probability Distribution: -brainly.com/question/9333901

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6 0

1 year ago

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the resulting values to the corresponding l

belka [17]

The correct solution to the limits of x in the tiles can be seen below.

$\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }$ $\mathbf{ = -\dfrac{1}{52} }$

$\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }$ $\mathbf{=\dfrac{1}{79} }$

$\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }$ = -31

$\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }$ $\mathbf{ =\dfrac{1}{98}}$

<h3>What are the corresponding limits of x?</h3>

The limits of x approaching a given number of a quadratic equation can be determined by knowing the value of x at that given number and substituting the value of x into the quadratic equation.

From the given diagram, we have:

$\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }$

So, x - 9 is positive when x → 9⁺. Therefore, |x -9) = x - 9

$\mathbf{ \lim_{x \to 9^+} (\dfrac{x-9}{-x^2-34+387}) }$

Simplifying the quadratic equation, we have:

$\mathbf{ \lim_{x \to 9^+} (-\dfrac{1}{x+43}) }$

Replacing the value of x = 9

$\mathbf{ = (-\dfrac{1}{9+43}) }$

$\mathbf{ = -\dfrac{1}{52} }$

$\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }$

-x²-63x+568 is positive when x → 8⁻.

Thus |-x²-63x+568| = -x²-63x+568

$\mathbf{ \lim_{x \to 8^-} (\dfrac{1}{x+71}) }$

$\mathbf{=\dfrac{1}{8+71} }$

$\mathbf{=\dfrac{1}{79} }$

$\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }$

x -7 is positive, therefore |x-7| = x - 7

$\mathbf{ \lim_{x \to 7^+} (\dfrac{-x^2-17x+168 }{x-7}) }$

$\mathbf{ \lim_{x \to 7^+} (-x-24)}$

$\mathbf{ \lim_{x \to 7^+} (-7-24)}$

= -31

$\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }$

x-6 is negative when x → 6⁻. Therefore, |x-6| = -x + 6

$\mathbf{ \lim_{x \to 6^-} (\dfrac{-x+6 }{-x^2-86x+552}) }$

$\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{x+92}) }$

$\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{6+92}) }$

$\mathbf{ =\dfrac{1}{98}}$

Learn more about calculating the limits of x here:

brainly.com/question/1444047

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7 0

2 years ago

Solve for s.<br> S + 3s - 4 = 16<br> S =<br> Submit

const2013 [10]

The answer is 5 because 16+4=20 and 3s + s = 4s so you’re left with 4s = 20 and I know that 4 times 5 is 20.
ANSWER: S=5

3 0

3 years ago

write and show on a number line the set of all the numbers that are less than 0 and greater than (-5)

vodka [1.7K]

Answer:

0 i think................??????

4 0

3 years ago

Jared drew this rectangle. Then, he drew another rectangle that was twice as long but had the same width. In what way do the are

kenny6666 [7]

Answer:

Step-by-step explanation:

one rectangle has the area, A₁=LxW and,

perimeter P₁=2L+2W

the second rectangle has the area A₂=2L x W,

and perimeter P₂=2(2L)x2W= 4Lx2W

the second area doubles compared to the first area

the perimeter gets increased with double the length

3 0

3 years ago