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gtnhenbr [62]
1 year ago
11

Amir is operating a machine that adds sand to a concrete mixture. The machine adds 1/10 ton of sand to the mixture in 4/5 minute

. Enter a number in each box to complete the statements. Write the ratio that compares 1/10 ton of sand to 4/5 minute. The ratio in ton(s) per minute(s) is $$ ​. Simplify the ratio to find the amount of sand added in one minute. ton(s) per minute
Mathematics
1 answer:
zavuch27 [327]1 year ago
7 0

for the above question, we can first convert the fractions to decimals to make it easier to work with

<span>1/10 of sand - 0.1 ton </span>

4/5 minute - 0.8 minute

<span>Q1 - </span>

<span>ratio </span>

0.1:0.8

since the ratio is in decimals we can convert it to whole numbers by multiplying by 10

1:8

Q2. amount of sand added in 0.8 minutes - 0.1 ton

<span>       amount of sand added in 1 minute - 0.1/0.8 = 0.125 tons</span>

<span>
</span>
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5. You deposit $300 in an account Urat pays 1.48% annual interest. What is the balance after 1 year if the
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The correct option is (A) $304.47.

Step-by-step explanation:

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FV=A\times [1+\frac{r\%}{365}]^{n\times 365}

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Recent homebuyers from a local developer allege that 30% of the houses this developer constructs have some major defect that wil
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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

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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

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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
1 year ago
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