The sum of three numbers is 8. The third is 9

Mathematics

1 answer:

Ierofanga [76]3 years ago

7 0

Answer:

→ First value is 1

→ Second value is 2

→ Third value is 5

Step-by-step explanation:

• let numbers be x, y and z

${ \tt{z = 8y - 9 - - - (eqn \: 1)}} \\ \\ { \tt{10x = 8y - 7 - - - (eqn \: 2)}} \\ \\ { \tt{x + y + z = 8 - - - (eqn \: 3)}}$

• from eqn 2, make x the subject:

${ \tt{x = \frac{8y - 7}{10} }} \\$

• substitute all variables in eqn 3:

${ \tt{ \frac{8y - 7}{10} + y + 8y - 9 = 8}} \\ \\ { \tt{8y - 7 + 10y + 80y - 90 = 80}} \\ \\ { \tt{98y = 177}} \\ \\ { \boxed{ \tt{ \: y = 1.8}}}$

• find z

${ \tt{z = 8y - 9}} \\ \\ { \tt{z = 8(1.8) - 9}} \\ \\ { \tt{z = 14.4 - 9}} \\ \\ { \boxed{ \tt{ \: z = 5.4 \: }}}$

• find x:

${ \tt{x = \frac{8(1.8) - 7}{10} }} \\ \\ { \tt{x = \frac{7.4}{10} }} \\ \\ { \boxed{ \tt{ \: x = 0.74 \: }}}$

Rounding to nearest value:

${ \boxed{ \rm{x = 1}}} \\ { \boxed{ \rm{y =2 }}} \\ { \boxed{ \rm{z = 5}}}$

You might be interested in

"A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean

Arada [10]

Answer:

85.31% probability that their mean rebuild time exceeds 8.1 hours.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$