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3 years ago

1. The absolute value of |-8|= ?

1 answer:

7 0

1. Absolute value is the distance that a number is away from 0. Hence, the answer is 8.

2. Same thing for this one: 14

3. -11, -3, 0, 7, 10. The bigger a negative number, the less it is. The bigger a positive number is, the greater it is.

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If the life in years of a television set is normally distributed with a mean of 36 years and a standard deviation of 8 years wha

TiliK225 [7]

Answer:

The company should guarantee a lifetime of less than equal to 20.95 years so that less than 3% of the television sets fail while under warranty.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 36 years

Standard Deviation, σ = 8 years

We are given that the distribution of life of television sets is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.03.

$P( X < x) = P( z < \displaystyle\frac{x - 36}{8})=0.03$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 36}{8} =-1.881\\\\x = 20.952\approx 20.95$

Thus, the company should guarantee a lifetime of less than or equal to 20.95 years so that less than 3% of the television sets fail while under warranty.

3 0

3 years ago

2sin^2 2x=2<br><br> In the domain [0,2pi)<br><br> What can x equal ( in radians).

lions [1.4K]

Bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,

$\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2}
\\\\\\
sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1}
\\\\\\
sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)$

$\bf \measuredangle 2x=sin^{-1}(\pm 1)\implies \measuredangle 2x=
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}\\\\
-------------------------------\\\\
\measuredangle 2x=\cfrac{\pi }{2}\implies \measuredangle x=\cfrac{\pi }{4}\qquad \qquad \qquad \qquad \measuredangle 2x=\cfrac{3\pi }{2}\implies \measuredangle x=\cfrac{3\pi }{4}$

3 0

3 years ago

Evaluate the line integral, where c is the given curve. c y3 ds,

postnew [5]

Parameterizing $\mathcal C$ by

$\mathbf r(t)=(t^3,t)$

with $0\le t\le2$ , we have

$\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt=\sqrt{(3t^2)^2+1^2}\,\mathrm dt=\sqrt{9t^4+1}\,\mathrm dt$

So

$\displaystyle\int_{\mathcal C}y^3\,\mathrm ds=\int_{t=0}^{t=2}t^3\sqrt{9t^4+1}\,\mathrm dt$
$=\displaystyle\frac1{36}\int_0^236t^3\sqrt{9t^4+1}\,\mathrm dt$
$=\displaystyle\frac1{36}\int_{u=1}^{u=145}\sqrt u\,\mathrm du$
$=\dfrac{145^{3/2}-1}{54}$

5 0

4 years ago

Is (1, -2) a solution of the graphed inequality?

jeyben [28]

Answer:

No, it is not it is a coordinate point

A graphed inequality x<8 that example

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Given the functions, f(x) = x2 + 2 and g(x) = 4x - 1, perform the indicated operation. When applicable, state the domain restric