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3 years ago

One number is 9 less than the other. Find the two numbers if their product is 736.

1 answer:

6 0

$x-first\ number\\x-9-second\ number\\\\x(x-9)=736\\x^2-9x-736=0\\\\\Delta=b^2-4ac\\\\\Delta=(-9)^2-4\cdot(-736)\cdot1=81+2944=3025\\\\\sqrt\Delta=55\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\ \ x_1=\frac{9-55}{2*1}=\frac{-46}{2}=-23
\\\\ or \\
x_2=\frac{-b+\sqrt{\Delta}}{2a}\ \ x_2=\frac{9+55}{2*1}=\frac{64}{2}=32\\\\x_1\ is\ negative\ number\ so\ it\ can't\ be\ one\ of\ them\\\\32-9=23\\\\Searched\ number\ are\ 32\ and\ 23
$

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Answer: 1/24

Step-by-step explanation:

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3 years ago

If f(x) = -3x - 5 and g(x) = 4x - 2, find (f+ g)(x).

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Answer:

$\boxed{\sf x - 7}$

Given:

f(x) = - 3x - 5

g(x) = 4x - 2

To Find:

(f + g)(x)

Step-by-step explanation:

$\sf (f + g)(x) = f(x) + g(x) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 3x - 5 + 4x - 2 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4x - 3x - 5 - 2 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = x - 5 - 2 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = x - 7$

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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.