Y=40+5x
For an example, say Mia wanted to add $5 into her account each week for 2 weeks.
y=40+5(2)
5(2)=10
y=40+10
y=50
Mia will have a total of $50 in her account if she added $5 for each week for 2 weeks. Same for Juan, his equation will be based off his amount y=15+10x

4 0

2 years ago

A transformation maps a onto b and c onto d name the image of a name the preimage of b name the image of c name the preimage of

Doss [256]

If a transformation maps $a \mapsto b$ , then $b$ is the image of $a$ , and $a$ is the preimage of $b$ . So,

the image of $a$ is $b$
the preimage of $b$ is $a$
the image of $c$ is $d$
the preimage of $d$ is $c$

7 0

2 years ago

Whats the answer to this numeric algebraic expression

Degger [83]

Answer:

a+b+2

Step-by-step explanation:

4 0

2 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 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 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.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

6 0

2 years ago