What is the x-coordinate of the x-intercept of the equation 4y + 5x + 13 = 3x +9y

Please help will mark you brainliest. ( all links will be reported)

cestrela7 [59]

Answer:

$6.22

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

NEED HELP FAST PLEASE

natita [175]

Hey!

To solve this equation, we'd first have to apply the distributing rule to this equation.

Original Equation :
$( 3a - 2 )^{2}$

New Equation {Changed by Applying the Distribution Rule} :
$( 3a )^{2} - 2$ · 3a · 2 + $2 ^{2}$

And now we simplify.

Old Equation :
$( 3a ) ^{2} -2$ · 3a · 2 + $2 ^{2}$

New Equation {Old Equation Simplified} :
$9a ^{2}$ - 12a + 4

So, the equation $(3a - 2) ^{2}$ simplified is $9a ^{2}$ - 12a + 4.

Hope this helps!

- Lindsey Frazier ♥

5 0

3 years ago

4. Order the numbers from least to greatest. √5, 5, 2.5. Ja

lara31 [8.8K]

Answer:

√5 < 2.5 < 5

Step-by-step explanation:

√5= 2.23

therefore,

√5 < 2.5 < 5

8 0

4 years ago

The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.

Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

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.

If $Z \leq -2$ or $Z \geq 2$ , the outcome X is considered unusual.

Central Limit Theorem

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