Solve for B<br><br> b

If Nathan scoops 8.5 cups of ice cream in 2 minutes, how many scoops of ice cream can he do in 8 minutes?

lozanna [386]

Answer:

34

Step-by-step explanation:

in one minute he scoops 4.25 cups which you get by dividing 8.5 and 2 and then you multiply that by 8 and you get 34 cups

8 0

3 years ago

Read 2 more answers

For brainiest:):):):):):):)

andrew-mc [135]

Answer:

15 = 0.025

16= 3 years

Step-by-step explanation:

15.

I=prt

15= (200)(r)(3)

200(3) = 600r

15 divided by 600r = 40 (0.025)

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16.

I=prt

$t= \frac{I}{Ixp} \\t= \frac{90}{0.05x600} \\t= 3$

t= 3 years

5 0

3 years ago

Determine if the following scenario is inductive or deductive reasoning:

Elan Coil [88]

You guys should go eat and then come pick ya guys up and then go pick ya up and go pick ya up some stuff and go get some stuff to go pick up some lunch wyeywywywyywywywyeh was your day so you can come pick ya up and then come pick me up I wanna is the day I wanna come to hang out with you guys tomorrow night or tomorrow or Saturday or tomorrow night or tomorrow or Saturday or Sunday or Saturday or tomorrow or Saturday or Sunday or Saturday night or Sunday or 22

8 0

3 years ago

A radio station claims that the amount of advertising per hour of broadcast time has an average of 13 minutes and a standard dev

zimovet [89]

Answer:

$Z = 3.33$

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu = 13, \sigma = 1.2$

You listen to the radio station for 1 hour, at a randomly selected time, and carefully observe that the amount of advertising time is equal to 17 minutes. Calculate the z-score for this amount of advertising time.

We have to find Z when X = 17. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17 - 13}{1.2}$

$Z = 3.33$

5 0

3 years ago

A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit

Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = - (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] = 20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

7 0

3 years ago

Solve for B b - 15 = a