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goldenfox [79]
2 years ago
7

What number is between 0.40 and 0.47 it has to be a fraction

Mathematics
2 answers:
Mademuasel [1]2 years ago
5 0
There are an infinite number of possibilities, but one of them could be 45/100 or 9/20
Lady_Fox [76]2 years ago
5 0
23/50 you take for example 46, put it over 100 which is 46/100 and then reduce.

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Which is bigger 0.96 0r 19/20
REY [17]
0.96 or 19/20

0.96 as fraction = 96/100 = 19.2 / 20      5 into 96 is 19.2 and 5 into 100 is 20

So 0.96 is the same as 19.2/20

Since 19.2 / 20  and 19 / 20 have the same denominator,  therefore the 19.2 / 20 is bigger than 19 / 20.

So the 0.96 is bigger than the 19 / 20.
7 0
2 years ago
The approximate areas of two states are listed below. Texas: 2.69\times10^5 square miles Rhode Island: 1.21\times10^3 square mil
Pavlova-9 [17]

Answer:

2.6779*10^5

Step-by-step explanation:

The problem given is 2.69*10^5 - 1.21*10^3 which is 269000-1210=267790 -> 2.6779*10^5

3 0
1 year ago
Read 2 more answers
Match each set of prices with the correct percent of increase or decrease.
cricket20 [7]
<span>Take the
(original price - new price)/original price
then multiply by 100
If it is a positive number, it is a decrease.
If it is a negative number, it is an increase.

original price: $63.00
new price: $56.07
11% decrease

</span><span>original price: $12.25
new price: $13.23
</span><span>8% increase
</span>
<span>original price: $98.00
new price: $108.78
</span>11% increase

<span>original price: $37.50
new price: $34.50
8% decrease</span>
7 0
1 year ago
Suppose you are told that a distribution is said to be approximately normal because outliers have skewed the data. List the poss
andrew-mc [135]

Answer:

Step-by-step explanation:

Outline are values which are entirely different from those remaining values in a data set. These extreme values can skew an approximately normal distribution by skewing the distribution in the direction of the outliers and this makes it difficult for the data set to be analyzed.

Its effect is such that the mean becomes extremely sensitive to extreme outliers making it possible that the mean is this not a representative of the population and this theoretically affects the standard deviation.

8 0
2 years ago
Recent homebuyers from a local developer allege that 30% of the houses this developer constructs have some major defect that wil
umka21 [38]

Answer:

3.54% probability of observing at most two defective homes out of a random sample of 20

Step-by-step explanation:

For each house that this developer constructs, there are only two possible outcomes. Either there are some major defect that will require substantial repairs, or there is not. The probability of a house having some major defect that will require substantial repairs is independent of other houses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

30% of the houses this developer constructs have some major defect that will require substantial repairs.

This means that p = 0.3

If the allegation is correct, what is the probability of observing at most two defective homes out of a random sample of 20

This is P(X \leq 2) when n = 20. So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.3)^{0}.(0.7)^{20} = 0.0008

P(X = 1) = C_{20,1}.(0.3)^{1}.(0.7)^{19} = 0.0068

P(X = 2) = C_{20,2}.(0.3)^{2}.(0.7)^{18} = 0.0278

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0008 + 0.0068 + 0.0278 = 0.0354

3.54% probability of observing at most two defective homes out of a random sample of 20

5 0
2 years ago
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