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Anna [14]
3 years ago
9

HELP! Suppose you make hot chocolate. It's too hot to drink, so you put it in the freezer to cool it off. If the original temper

ature is 190 degrees Fahrenheit and the freezer is 0 degrees Fahrenheit, what will the temperature of the hot chocolate be after 4 minutes? (k = 0.12)
Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

go follow this website it has a full answer brainly.com/question/2803975

Step-by-step explanation:

You might be interested in
Question 5
geniusboy [140]

Answer:

\frac{4}{9}

Step-by-step explanation:

To find this probability, we have to know:

  • How many numbers are there in between 5 and 8 inclusive?
  • How many total numbers are there?

We simply divide the first answer by the second one and get our probability.

So, the numbers are 5,6,7,8 ----- that is 4 numbers

How many numbers are there in total?  That is:

0, 1, 2, 3, 4, 5, 6, 7, 8, or 9  ------- 9 numbers

Thus, the probability is 4/9

7 0
3 years ago
Select the correct answer.
bezimeni [28]
I think it would be A
5 0
3 years ago
Food inspectors inspect samples of food products to see if they are safe. This can be thought of as a hypothesis test with the f
Novay_Z [31]

Answer:

Type II error

Step-by-step explanation:

According to the definition of type II error, the type II error arises when we wrongfully accept the null hypothesis. It means that when we accept our null hypothesis and null hypothesis is not correct then type II error arises. So according to situation we are accepting the null hypothesis that the food is safe but it is actually not safe. Hence the given situation represents type II error.

7 0
3 years ago
10. Two sides of a triangle have
Oksanka [162]
I think the answer is 6 from the information you provided.
7 0
3 years ago
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
3 years ago
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