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11Alexandr11 [23.1K]

3 years ago

14

How many minutes should she walk at a rate of 3 miles per hour to finish traveling the 4 miles?

1 answer:

SIZIF [17.4K]3 years ago

4 0

Answer:

She should walk for 12 minutes

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WILL GIVE BRAINLIEST

Montano1993 [528]

Step-by-step explanation:

Use the compound interest formula:

A = P( 1 + r/n)^nt

A is the amount at the end

P is the principal or original amount.

R is annual rate as a decimal (you have to convert)

N is the # of compounding periods in a year (quarterly means 4)

T is time in years

A = 500( 1 + 0.0475/4)^(4×20)

This would be $1285.64 when fully computed.

3 0

2 years ago

Solve for r 12-1/5r=2r+1

sveticcg [70]

Answer:

5r

Step-by-step explanation:

12−

1

5r

=

60r−1

5r

7 0

3 years ago

Complex root of x^2+3x+9=0

Luden [163]

$\text{Quadratic formula}$

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}$

-----------------------------------------------------------------------

$x^2+3x+9=0\\\\a=1;\ b=3;\ c=9\\\\\Delta=3^2-4(1)(9)=9-36=-27\\\\\sqrt\Delta=\sqrt{-27}=\sqrt{(27)(-1)}=\sqrt{27}\cdot\sqrt{-1}=\sqrt{9\cdot3}i=\sqrt9\cdot\sqrt3i=3\sqrt3i$

$x_1=\dfrac{-3-3\sqrt3i}{2(1)}=-\dfrac{3}{2}-\dfrac{3\sqrt3}{2}i\\\\x_2=\dfrac{-3+3\sqrt3i}{2(1)}=-\dfrac{3}{2}+\dfrac{3\sqrt3}{2}i$

$i=\sqrt{-1}$

6 0

3 years ago

Three lightship flash simultaneously at 6 00 a.m the first lightship flashes every 12 sec the second lightship every 30 sec and

Dmitrij [34]

They will all flash their lights together at 5:00 pm

6 0

2 years ago

One hundred items are simultaneously put on a life test. Suppose the lifetimes

romanna [79]

Answer:

a) $\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

b) $\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( $i-1$ )st and the $ith$ failures. Then, the $T_{i}$ are independent with $\mathrm{T}_{\mathrm{i}}$ being exponential with rate $\frac{(101-i)}{200} .$

Therefore,

a) $E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]$

$=\sum_{i=1}^{5} \frac{200}{101-i}$

$\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

$b)$

The variance is given by, $\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]$

$\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

7 0

3 years ago