Write a decimal that is less than 6.73

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

7 0

5.2 is less because it's a lower number and still a decimal

vaieri [72.5K]3 years ago

3 0

Anything that has a decimal place and is less than 6.73 would work. 6.72, 5.89, anything that is not 6.73 or greater than it.

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romanna [79]

Answer:

The probability you will get a head at least once is 50%.

Step-by-step explanation:

Since the question is asking about the probability you will get, we can assume we’re answering based on theoretical probability. This type of probability is based on logic.

A coin always has two sides, one with head and the other with tails. So we can easily represent this as half and half. 1/2 as a fraction. 0.5 as a decimal. 50% as a percent. This means that P(H) will be equal to any one of these as they are all the same. The same can be said for the probability that a head does not appear, in other words, a tail appears. The reason being that the probability is split evenly between the two. This will again mean that P(T) will equal to any one of those.

So, A = 50% and B = 50%. The probability you will get a head at least once is 50%.

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3 years ago

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sergey [27]

Answer:

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Step-by-step explanation:

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4 0

2 years ago

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: 3 tires are blemished.

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: 2 tires are blemished.

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: 1 tires are blemished.

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: No tires are blemished.

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .