The diagram shows the straight line which passes through the points (0, 1) and (3, 13).

Mathematics

1 answer:

Damm [24]2 years ago

5 0

Answer:

y=4x+1

Step-by-step explanation:

First find the gradient of the straight line

Then substitute the gradient in the formula y=mx+c.

Substitute one of the given coordinates in the line to find the value of c.

After finding the value of c substitute it in the formula but do not write the values of x and y in the formula

You might be interested in

Write an equation for the translation of x^2+y^2=25 by 7 units left and 2 units down

soldi70 [24.7K]

Your equation is a circle where the center is the origin (0, 0) and the radius is 5. With the translation, the new center is (- 7, - 2). Radius stays the same.

Your new equation is:

(x + 7)² + (y + 2)² = 25

4 0

3 years ago

A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of

Masteriza [31]

Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .