Please Help this is the only one i dont know!!

Mathematics

1 answer:

Stella [2.4K]3 years ago

7 0

Answer:

$\large\boxed{x=\dfrac{23}{33}}$

Step-by-step explanation:

$\text{Let}\ x=0.\overline{69}\\\\x=0.696969...\qquad\text{multiply both sides by 100}\\\\100x=69.696969...\qquad\text{subtract}\ x\ \text{from}\ 100x:\\\\100x-x=69.696969...-0.696969...\\\\99x=69\qquad\text{divide both sides by 99}\\\\x=\dfrac{69}{99}\\\\x=\dfrac{23}{33}$

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W=zr-2zt^3 solve for z

ladessa [460]

Answer: z=0

Step-by-step explanation:

W=zr-2zt^3

Move all terms containing z to the left, all other terms to the right.

dd '-1rz' to each side of the equation.

-1rz+z=rz+-1rz+-2t^3z

Combine like terms: rz + -1rz = 0

-1rz+z=0+(-2t^3z)

-1rz+z=-2t^3z

Add 2t^3z to each side

-1rz+2t^3z+z=-2t^3z+2t^3z

Combine the like terms 2t^3z+2t^3z=0

-1rz+2t^3z+z=0

Factor out the Greatest Common Factor (GCF), 'z'.

z(-1r+2t^3+1)=0

Hope this helps, HAVE A BLESSED AND WONDERFUL DAY! As well as a great Superbowl Weekend! :-)

- Cutiepatutie ☺❀❤

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3 years ago

The body temperatures of adults are normally distributed with a mean of 98.6degrees° F and a standard deviation of 0.60degrees°

Schach [20]

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples can be 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 $s = \frac{\sigma}{\sqrt{n}}$