Just look at the picture

Please help me with a problem of Derive the Formula for Area of a Triangle

ss7ja [257]

You can use
$\frac{1}{2} ab \sin(c)$
Since it is an irregular triangle.
C refers to the angle and a & b refer to the adjacent sides.

So,
$\frac{1}{2} \times4.3 \times 3.8 \sin(53) \\ = 6.52485211 \\ = 6.52$

4 0

3 years ago

Which Expression Can Be Used To Determine the nth term in the pattern below<br> 20,25,30,35,40

maw [93]

Answer:

see explanation

Step-by-step explanation:

These are the terms of an arithmetic sequence with n th term

$a_{n}$ = a + (n - 1)d

where a is the first term and d the common difference

d = 25 - 20 = 30 - 25 = 5 and a = 20, hence

$a_{n}$ = 20 + 5(n - 1) = 20 + 5n - 5 = 5n + 15 ← n th term formula

8 0

3 years ago

The standard deviation of the scores on a skill evaluation test is 260 points with a mean of 1612 points. If 344 tests are sampl

shepuryov [24]

Answer:

3.24% probability that the mean of the sample would differ from the population mean by greater than 30 points

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 1612, \sigma = 260, n = 344, s = \frac{260}{\sqrt{344}} = 14.02[qtex]What is the probability that the mean of the sample would differ from the population mean by greater than 30 pointsEither it differs by 30 or less points, or it differs by more than 30 points.The sum of the probabilities of these events is 100. So i will first find the probability that it differs by 30 or less points.Probability that it differs by 30 or less pointsPvalue of Z when X = 1612 + 30 = 1642 subtracted by the pvalue of Z when X = 1612 - 30 = 1582. SoX = 1642[tex]Z = \frac{X - \mu}{\sigma}$