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2 years ago

133 1/3 % of what number is 74?

Mathematics

2 answers:

alexandr402 [8]2 years ago

5 0

Answer:

The number is 55.5

Step-by-step explanation:

Consider the provided information.

Let the number is x.

It is given that 133 1/3 % of x is 74.

Now convert the above statement into mathematical form.

$133\frac{1}{3}\% \times x =74$

$\frac{400}{3}\times \frac{1}{100} \times x =74$

$\frac{4}{3}\times x =74$

$x =74\times \frac{3}{4}$

$x =18.5\times 3$

$x =55.5$

Hence, the number is 55.5

vfiekz [6]2 years ago

4 0

1.333 ×x = 74

x =
$\frac{74}{1.333} = 55.50$

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Determine whether the sequence is arithmetic or geometric.

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Answer:

Sequence 1 is arithmetic, sequence 2 is geometric.

Step-by-step explanation:

Sequence 1 is being subtracted by 16 each time, so it is arithmetic.

Sequence 2 is being divided by 2 each time, so it is geometric.

Both, sequence 1 and 2, have terms 32 and 16.

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2 years ago

The probability that an Oxnard University student is carrying a backpack is .70. If 10 students are observed at random, what is

Rus_ich [418]

Answer:

35.03% probability that fewer than 7 will be carrying backpacks

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are carrying a backpack, or they are not. The probability of a student carrying a backpack is independent from other students. So we use the binomial probability distribution to solve this question.

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

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.

The probability that an Oxnard University student is carrying a backpack is .70.

This means that $p = 0.7$

If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks

This is $P(X < 7)$ when $n = 10$ . So

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

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$P(X = 3) = C_{10,3}.(0.7)^{3}.(0.3)^{7} = 0.0090$

$P(X = 4) = C_{10,4}.(0.7)^{4}.(0.3)^{6} = 0.0368$

$P(X = 5) = C_{10,5}.(0.7)^{5}.(0.3)^{5} = 0.1029$

$P(X = 6) = C_{10,6}.(0.7)^{6}.(0.3)^{4} = 0.2001$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.000006 + 0.0001 + 0.0014 + 0.0090 + 0.0368 + 0.1029 + 0.2001 = 0.3503$

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2 years ago

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kondor19780726 [428]

Answer:

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Step-by-step explanation:

The parameter given are;

To 12 × 15 by changing the numbers into friendlier numbers

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