How can I solve this? x= AD= AB= 5X-4 DB=x+1

6TH GRADE MATH. PLEASE HELP! 25 POINTS!

Helen [10]

Answer:

Express as exponents) 1) 81 / 2) 4/9 3) -64

Challange: Use parenthesis) 1) 3b 2) 8x^3y^3 3)15x^2y^2

Evaluate) 1) 125 2) 81/256

Challenge) 9a^2b^2

True or flase) 1) flase 2) false 3) true 4) false

Step-by-step explanation:

6 0

3 years ago

Consider the probability that no less than 95 out of 152 registered voters will vote in the presidential election. Assume the pr

nikdorinn [45]

Answer:

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given registered voter will vote in the presidential election is 61%.

This means that $p = 0.61$

152 registed voters:

This means that $n = 152$

Mean and Standard deviation:

$\mu = E(X) = 152*0.61 = 92.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{152*0.61*0.39} = 6.01$

Probability that no less than 95 out of 152 registered voters will vote in the presidential election.

This is, using continuity correction, $P(X \geq 95 - 0.5) = P(X \geq 94.5)$ , which is 1 subtracted by the pvalue of Z when X = 94.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{94.5 - 92.72}{6.01}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

6 0

3 years ago

The population standard deviation for the temperature of beers found in Scooter's Tavern is 0.26 degrees. If we want to be 90% c

siniylev [52]

Answer:

19 beers must be sampled.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The population standard deviation for the temperature of beers found in Scooter's Tavern is 0.26 degrees.

This means that $\sigma = 0.26$

If we want to be 90% confident that the sample mean beer temperature is within 0.1 degrees of the true mean temperature, how many beers must we sample?

This is n for which M = 0.1. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.1 = 1.645\frac{0.26}{\sqrt{n}}$