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Stells [14]
3 years ago
13

What is the greatest possible product of a 2 digit number and a 1 digit number?

Mathematics
2 answers:
Law Incorporation [45]3 years ago
6 0
Largest 1-digit integer: 9
Largest 2-digit integer: 99
9 * 99 = 891
sladkih [1.3K]3 years ago
5 0
Highest 2-digit number = 99 
<span>Highest 1-digit number = 9 </span>
<span>Highest possible sum from 2, 2-digit numbers = 198 </span>
<span>Highest possible sum from 2, 1-digit numbers = 18</span>
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A Walmart truck driver estimates that it will take him 12 hours to drive 1152 km. After 5 hours, he has driven 520 km.
san4es73 [151]

(a) He would drive 96 km in an hour because 1152 divided by 12 is 96

(b) He is technically on schedule because he is driving 104 km per hour, instead of 96, so if he keeps driving at the same rate, he should get there in under 12 hours

6 0
3 years ago
If the area of a rectangle is 144 sq. in. and the length of one side is 18in. what is the length of the other side?
navik [9.2K]

Hello from MrBillDoesMath!

Answer:  The other side has a length of 8

Discussion:

Area of a rectangle = l * w ( length times width). In our case

Area = 144 = l * w = 18 * w =>

18 w = 144  =>

w = 144/18 = 8


Thank you,

MrB

8 0
3 years ago
Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo
natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

So

P(B) is the probability of the person having cancer, so P(B) = 0.009

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So P(A/B) = 0.91

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765

What is the probability that the person actually does have cancer?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213

There is a 12.13% probability that the person actually does have cancer.

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