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3 years ago

30 POINTS TO WHOEVER HELPS WITH THIS I NEED IT ASAP PLEASE HELP

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

5 0

Answer:

4a ) Difference ⇒ $ 63.24,

4b ) 4 children

Step-by-step explanation:

4a ) If the number of children is represented by the value of x, let us substitute a family of 2 children as value of x into this equation we are given;

y = 21.08 * ( 2 ) + 85.15 ⇒ Multiply 21.08 by 2,

y = 42.16 + 85.15 ⇒ Combine like terms,

y = 127.31

If y ⇒ amount spent on grocery's, let us solve for the amount spent on 5 children, and subtract from this amount;

y = 21.08 * ( 5 ) + 85.15 ⇒ Multiply 21.08 by 5,

y = 105.4 + 85.15 ⇒ Combine like terms,

y = 190.55

Difference; 190.55 - 127.31 = 63.24,

Solution; Difference ⇒ $ 63.24

4b ) If the amount is $ 169.47, let us plug it into our given equation provided amount spent on groceries ⇒ y;

169.47 = 21.08x + 85.15 ⇒ Subtract 85.15 from either side of equation,

21.08x = 84.32 ⇒ Divide either side 21.08,

x = 4 children

Solution; 4 children

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3 years ago

Read 2 more answers

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.