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d1i1m1o1n [39]
2 years ago
14

Please can you give me the answers to the questions above in standard form.​

Mathematics
1 answer:
ivanzaharov [21]2 years ago
3 0

Answer:

a) 4.5 \times 10^{11}

b) 3.5 \times 10^{3}

Step-by-step explanation:

a) We have to simplify the below expression in standard form.

The expression is (5\times 10^{3}) \times (9 \times 10^{7})

Now, (5\times 10^{3}) \times (9 \times 10^{7})

= (9 \times 5) \times (10^{3} \times 10^{7})

= 45 \times 10^{(3 + 7)} {Since a^{b} \times a^{c} = a^{(b + c)}}

= 45 \times 10^{10}

= 4.5 \times 10^{11} (Answer)

b)  We have to simplify the below expression in standard form.

The expression is (7\times 10^{5}) \div (2 \times 10^{2})

Now, (7\times 10^{5}) \div (2 \times 10^{2})

= (7 \div 2) \times (10^{5} \div 10^{2})

= 3.5 \times 10^{(5 - 2)} {Since a^{b}\div a^{c} = a^{(b - c)}}

= 3.5 \times 10^{3} (Answer)

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A financial talk show host claims to have a 55.3 % success rate in his investment recommendations. You collect some data over th
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Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

p = 0.553, n = 10

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

In which

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

P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003

P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039

P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219

P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724

P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

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The inference that can be made about the results is that; energy drinks enhances performance during 100 meters race.

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From the given table and histogram, we can see that;

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Read more about Inference at; brainly.com/question/1611703

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