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2 years ago

Which expression is less than 3? 3 x 4/4, 3 x 15/8, 3 x 4/9, 3 x 1 1/8

Mathematics

1 answer:

pav-90 [236]2 years ago

8 0

Answer:

3 × 4/9

Step-by-step explanation:

Remember that 4/4 is equal to 1, so 3 × 4/4 = 3

15 is bigger than 8, so we know that 15/8 is equal to 1 point something, you can find out the answer by a simple division, once you know the answer (1.875), you multiply it by 3, and get 5.625, this is larger than 3 so it is not the answer.

4 is smaller than 9, so we know that 4/9 is equal to 0 point something, again, we divide and get 0.444..., now we know that multypling something by 0 pint something is almost the same as dividing, so we can that this result is less than 3 (its 1.333).

1 and 1/8 is a mixed fraction so we know it is larger than 1, meaning that if we multiply it by 3, we will get a number larger than 3 (3.375)

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2 years ago

18, 26, 34, 42, find the 52nd term and the explicit formula

daser333 [38]

We have 26 - 18 = 8; 34 - 26 = 8; 42 - 34 = 8; (an arithmetic progression)
We use the formula an = a1 + (n-1)x8 (the explicit formula)
So, an = 18 + (n-1)x8 ;
Then, an = 8n + 10;
Finally, a52 = 8x52 + 10 = 426 (the 52nd term);

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3 years ago

Can someone please help me with this question?

scoray [572]

The answer is the first one

7 0

3 years ago

An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0

zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 105, p = 0.9$

$\mu = E(X) = np = 105*0.9 = 94.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.9*0.1} = 3.07$

What is the probability that everyone who appears for the flight will get a seat

100 or less people appearing to the flight, which is the pvalue of Z when X = 100. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{100 - 94.5}{3.07}$

$Z = 1.79$

$Z = 1.79$ has a pvalue of 0.9633

96.33% probability that everyone who appears for the flight will get a seat

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2 years ago

A candy store grab bag contains 10 pieces of sour candy, 12 pieces of chocolate candy, and 6 pieces of sweet candy. If you selec

harkovskaia [24]

Step-by-step explanation:

so we're making two draws *with* replacement (this is important)

step 1: for the first draw, it wants the probability of getting a sour candy. to calculate this:

(# of sour candy) / (total # of candy)

step 2: for the second draw, it wants the probability of *not* getting a sour candy. to calculate this, you can calculate 1 - (the probability form part 1).

step 3: to find the probability of both events happening together, simply multiply the probabilities from part 1 and 2 together

side note: for step 2, you can only do this because the candy is being replaced. if there were no replacement, you'd have to re-calculate (# of non-sour candies) / (total after the first candy is drawn)

4 0

2 years ago

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