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2 years ago

Kelly runs a distance of 100 metres in a time of 10.52 seconds.

1 answer:

7 0

The upper bound means the highest number that you can round up to the nearest unit, in this case meters. So the upper bound for the distance of 100 meters is 100.4 meters.

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Help me pleaseeeeeee

Pepsi [2]

so I don't know but just use the internet

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2 years ago

Marie has renters insurance that she must pay twice a year. If each payment is $96, how much money should she set aside each mon

Vlad [161]

Answer:

The answer is <u>B.) $16</u>

Step-by-step explanation:

Multiply 16$ by 6 which resembles half a year. You will get $96 from 16*6

4 0

1 year ago

Please help. i don't understand

zaharov [31]

78×.33=25.74 busses are new

25.74+15=40.74 new busses

78+15=93 total busses

40.74/78=0.5223076923= 52% will be new with the new 15 busses

4 0

2 years ago

Can anyone help me?

svp [43]

Answer:

Literally take your f(x) and - your g(x)

(x^2+9x+18) - (x + 6)

Remember to distribute the negative sign.

6 0

2 years ago

The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she d

aksik [14]

Answer:

The number of minutes advertisement should use is found.

x ≅ 12 mins

Step-by-step explanation:

(MISSING PART OF THE QUESTION: AVERAGE WAITING TIME = 2.5 MINUTES)

For such problems, we can use probability density function, in which probability is found out by taking integral of a function across an interval.

Probability Density Function is given by:

$f(t)=\left \{ {{0 ,\-t$

Consider the second function:

$f(t)=\frac{e^{-t/\mu}}{\mu}\\$

Where Average waiting time = μ = 2.5

The function f(t) becomes

$f(t)=0.4e^{-0.4t}$

The manager wants to give free hamburgers to only 1% of her costumers, which means that probability of a costumer getting a free hamburger is 0.01

The probability that a costumer has to wait for more than x minutes is:

$\int\limits^\infty_x {f(t)} \, dt= \int\limits^\infty_x {}0.4e^{-0.4t}dt$

which is equal to 0.01

Solve the equation for x

$\int\limits^{\infty}_x {0.4e^{-0.4t}} \, dt =0.01\\\\\frac{0.4e^{-0.4t}}{-0.4}=0.01\\\\-e^{-0.4t} |^\infty_x =0.01\\\\e^{-0.4x}=0.01$

Take natural log on both sides

$ln (e^{-0.4x})=ln(0.01)\\-0.4x=ln(0.01)\\-0.4x=-4.61\\x= 11.53$

<h3>Results</h3>

The costumer has to wait x = 11.53 mins ≅ 12 mins to get a free hamburger

3 0

2 years ago