Given f(x)=2\x^2-8x+36,find f(-4)

FRÉD AND ETHEL WACHED TV FOR SEVERAL HOURS ONE AFTERNOON. THEY BEGAN AT 1:30 P.M AND CONTINUED WACHING TV FOR 3 HOURS AND 45 MIN

QveST [7]

$30 + 45 =75$
$3 + 1 = 4$
There are 60 minutes in one hour, so 75 minutes is equal to 1 hour and 15 minutes.

$4 + 1 = 5$
Their mom turned the TV off at 5:15.

7 0

3 years ago

Evaluate x + y when x = -94 and y = 24 A: -118 B: -70 C: 70 D: 118

malfutka [58]

It’s b because 94 is negative so that means you have to subtract to get -70

8 0

3 years ago

Read 2 more answers

In college basketball, a shot made from beyond a designated arc radiating about 20 ft from the basket is worth three points ins

denis-greek [22]

Answer:

a) By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b) 4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered surprising, so yes, it would be surprising for him to make only 25 of these shots.

Step-by-step explanation:

To solve this question, we are going to 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 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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For proportions, we have that the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.45, n = 100$

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b)This is Z when X = 0.25.

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

By the Central Limit Theorem

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