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3 years ago

What is the answer to the long question?

Mathematics

1 answer:

gogolik [260]3 years ago

6 0

She added -14
the difference between 10 and -4 is 14
since this should use addition then we add -14

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The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000

valina [46]

Answer: 61,750 miles

Step-by-step explanation:

Given : The p-value of the tires to outlast the warranty = 0.96

The probability that corresponds to 0.96 from a Normal distribution table is 1.75.

Mean : $\mu=60,000\text{ miles}$

Standard deviation : $\sigma=1000\text{ miles}$

The formula for z-score is given by : -

$z=\dfrac{x-\mu}{\sigma}\\\\\Rightarrow\ 1.75=\dfrac{x-60000}{1000}\\\\\Rightarrow\ x-60000=1750\\\\\Rightarrow\ x=61750$

Hence, the tread life of tire should be 61,750 miles if they want 96% of the tires to outlast the warranty.

3 0

3 years ago

Evaluate the expression when x = 4 and y = 3. 2[(x - 1) + 2xy] A) 31 B) 40 C) 42 D) 54 E) 492

Dennis_Churaev [7]

When You Evaluate the expression you the Answer D) 54

7 0

3 years ago

Read 2 more answers

What is the mean, mode, median, and range of the numbers 24, 31, 12, 38, 12, and 15

tatyana61 [14]

Answer:

mean: 19.66667

mode: 12

median: 18

range: 37

8 0

3 years ago

A ship traveling east at 45 mph is 15 mi from a harbor when another ship leaves the harbor traveling east at 60 mph. How long do

Harman [31]

Answer:

1 hour

Step-by-step explanation:

I find these easiest to work by considering the initial difference in distance and the speed at with that gap is closing.

The gap is 15 miles, the distance the first ship is from harbor when the second ship starts.

The rate of closure is the difference in the speeds of the two ships:

60 mph -45 mph = 15 mph

Then the closure time is ...

time = distance/speed

time = (15 mi)/(15 mi/h) = 1 h

It will take the second ship 1 hour to catch up to the first ship.

7 0

2 years ago

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$

8 0

2 years ago