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4 years ago

you have 4 gallons of gas in your car. Your car gets 23 miles per gallon. How many miles can you drive?

Mathematics

2 answers:

attashe74 [19]4 years ago

7 0

You can drive 92 to miles and all you had to do to figure out the answer was 23x4

Ghella [55]4 years ago

4 0

Your car will take you 23 miles on the first gallon, 23 miles on the second gallon, 23 miles on the third gallon, and 23 miles on the fourth gallon. The total distance is 92 miles, and if you have not added more gas by that time, it will all stop right there.

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What does this mean? evaluate the expression for y=54y+10

salantis [7]

We have the expression:

$4y+10$

And want to evaluate in y=5, this means we have to replace y in the expression for 5 and calculate the value of the expression, so:

$\begin{gathered} 4y+10,y=5 \\ 4\cdot5+10 \\ 20+10 \\ 30 \end{gathered}$

The expression is 30 when y=5.

7 0

1 year ago

George earns $455 per week. George receives a 20% raise. How can George calculate his new weekly pay rate? Select all calculatio

Aleksandr [31]

It's only 455/x = 100 :))))))))

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3 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

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

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.