Please solve this problem

At 11 p.M., it was 48°F. The temperature was 6° cooler by 7 a.M. By 11 a.M. The temperature was 48°F again. What integer represe

Anvisha [2.4K]

Answer:

6°F

Step-by-step explanation:

Given that:

Temperature at 11pm = 48°F

Temperature at 7am = 6° cooler

Temperature at 11am = 48°F

Number of degrees temperature changed from 7am to 11 am

Exact temperature at 7am = (temperature at 11 pm - 6°)

= 48°F - 6°F

= 42°F

Temperature change from 7am to 11am

Temperature at 11 am - temperature at 7am

48°F - 42°F

= 6°F

6 0

3 years ago

A bacteria culture starts with 12,000 bacteria and the number doubles every 50 minutes.

Vlada [557]

Answer:

a) $y=12000(2)^{\frac{t}{50}}$

b) Approx. 27,569 bacteria

c) About 103 minutes

Step-by-step explanation:

a)

This will follow exponential modelling with form of equation shown below:

$y=Ab^{\frac{t}{n}}$

Where

A is the initial amount (here, 12000)

b is the growth factor (double, so growth factor is "2")

n is the number of minutes in which it doubles, so n = 50

Substituting, we get our formula:

$y=Ab^{\frac{t}{n}}\\y=12000(2)^{\frac{t}{50}}$

b)

To get number of bacteria after 1 hour, we have to plug in the time into "t" of the formula we wrote earlier.

Remember, t is in minutes, so

1 hour = 60 minutes

t = 60

Substituting, we get:

$y=12000(2)^{\frac{t}{50}}\\y=12000(2)^{\frac{60}{50}}\\y=12000(2)^{\frac{6}{5}}\\y=27,568.76$

The number of bacteria after 1 hour would approximate be <u>27,569 bacteria</u>

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c)

To get TIME to go to 50,000 bacteria, we will substitute 50,000 into "y" of the equation and solve the equation using natural logarithms to get t. Shown below:

$y=12000(2)^{\frac{t}{50}}\\50,000=12,000(2)^{\frac{t}{50}}\\4.17=2^{\frac{t}{50}}\\Ln(4.17)=Ln(2^{\frac{t}{50}})\\Ln(4.17)=\frac{t}{50}*Ln(2)\\\frac{t}{50}=\frac{Ln(4.17)}{Ln(2)}\\\frac{t}{50}=2.06\\t=103$