What is the value of z in the equation 4(2z + 3) = 12?

find the value of "a" and "b" for which the limit exists both as x approaches 1 and as x approaches 2:

lbvjy [14]

Answer:

a = 4

b = -2

Step-by-step explanation:

If the given function is continuous at x = 1

$\lim_{x \to 1^{-}} f(x)=(x+1)$

$=2$

$\lim_{x \to 1^{+}} f(x)=ax+b$

$=a+b$

$\lim_{x \to 1} f(x)=ax+b$

$=a+b$

And for the continuity of the function at x = 1,

$\lim_{x \to 1^{-}} f(x)=\lim_{x \to 1^{+}} f(x)=\lim_{x \to 1} f(x)$

Therefore, (a + b) = 2 -------(1)

If the function 'f' is continuous at x = 2,

$\lim_{x \to 2^{-}} f(x)=ax+b$

$=2a+b$

$\lim_{x \to 2^{+}} f(x)=3x$

$=6$

$\lim_{x \to 2} f(x)=3x$

$=6$

Therefore, $\lim_{x \to 2^{-}} f(x)=\lim_{x \to 2^{+}} f(x)=\lim_{x \to 2} f(x)$

2a + b = 6 -----(2)

Subtract equation (1) from (2),

(2a + b) - (a + b) = 6 - 2

a = 4

From equation (1),

4 + b = 2

b = -2

3 0

3 years ago

A manufacturer knows that their items have a normally distributed length, with a mean of 13.1 inches, and standard deviation of

Afina-wow [57]

Answer:

0.73% probability that their mean length is less than 11.1 inches

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 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$