-4log(2x) = -8<br> if someone can help me with this problem i’ll venmo them 2$

Figure ABCD is transformed to A′B′C′D′, as shown: A coordinate plane is shown. Figure ABCD has vertices A at 3 comma 1, B at 3 c

Marta_Voda [28]

4+4(7x5)= 7/6+58029+>667482+ªº993”QnQ no seeeeeeeeeeeeeeeeeeeeeeeeee
Eeeeeeeeeeeeeee mucho UwU
QwQ
Chauuuuu

7 0

3 years ago

Boden's account has a principal of $400 and a simple interest rate of 3.6%. Complete the number line. How much money will be i

Tju [1.3M]

400x1.036 then that multiplied by 1.036 and every answer just multiply by that

8 0

3 years ago

Read 2 more answers

Which expression is three less than the product of a number and two,increased by five

Tpy6a [65]

Answer:

(2x+5)-3

Step-by-step explanation:

6 0

2 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 $s = \frac{\sigma}{\sqrt{n}}$ .