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3 years ago

Jose works 30 hrs each week. At the end of 6 months (26weeks), he made $31,20. How much does Jose make per hour?

Mathematics

1 answer:

Ludmilka [50]3 years ago

6 0

Answer:

Jose makes $4 per hour

Step-by-step explanation:

1. 30x26=780

2. 3120 divide by 780=4

You might be interested in

Determine if the sequence below is arithmetic or geometric and determine the common difference / ratio in simplest form. 125, 25

Igoryamba

Answer:

Given sequence is an arithmetic sequence.

Step-by-step explanation:

Given sequence is:

125, 25, 5, .. ..

Here

$a_1 = 125\\a_2 = 25\\a_3 = 5$

A sequence is geometric sequence when the common ratio of consecutive terms is same. It is denoted by r.

A sequence is an arithmetic sequence when the common difference of consecutive terms is same. It is denoted by d.

Finding the common ratio

$r = \frac{a_2}{a_1} = \frac{125}{25} = 5\\r = \frac{a_3}{a_2} = \frac{25}{5} = 5$

As we can see that the common ratio is same, the given sequence is a geometric sequence.

5 0

3 years ago

Read 2 more answers

Oh, boy! Look at that monthly payment from Question 2 above! Molly cannot afford the monthly payment using the 0% financing. She

viktelen [127]

The car payment plan that has an APR of 1.9% is, on the short term, pocket

friendly.

The correct responses are;

First part: The amount of loan Molly needs is $22,495

Second part: Molly's monthly payment will be approximately $286.2

Third part: The total interest Molly will pay using the plan is approximately $1,545.8

Fourth part: The amount the car costs Molly using the plan is approximately $26,540.8

Reasons:

The question parameters are;

Molly is comparing Auto Loans on Jeep website

Selling price of the Jeep, P = $25,495

The down payment Molly has = $2,500

The Cash Allowance = $500

Number of months of payment, n = 84 months

The Annual Percentage Rate, APR, r = 1.9%

First Part:

Loan needed = $25,495 - $2,500 - $500 = $22,495

The amount of loan Molly needs = $22,495

Second Part:

The monthly payment is given by the formula;

$\displaystyle M = \mathbf{\dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}}$

Therefore;

$\displaystyle M = \dfrac{22,495 \times \left(\dfrac{0.019}{12} \right) \times \left(1+\dfrac{0.019}{12} \right)^{84} }{\left(1+\dfrac{0.019}{12} \right)^{84} - 1} \approx \mathbf{286.2}$

Molly's monthly payment will be M ≈ $286.2

Third part:

The total interest, I = Sum of payment - Loan amount

∴ The total interest, I ≈ $286.2/month × 84 month - $22,495 ≈ $1,545.8

The total interest Molly will pay using the loan, I ≈ $1,545.8

Fourth part:

The cost of the car, C, using the 1.9% APR financing plan is the sum of the down payment plus sum of the loan repayment

Therefore;

C ≈ $2,500 + $286.2/month × 84 months = $26,540.8

The cost of the car using the 1.9% APR financing plan, C ≈ $26,540.8

The question parameters obtained from a similar question are;

Selling price of the car = $25,495

Financing = 1.9% APR for 84 months with a cash allowance of $500

Learn more about loan payment here:

brainly.com/question/1040657

3 0

3 years ago

Please help with this question

amid [387]

Answer:

you like getting kinky

Step-by-step explanation:

6 0

3 years ago

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

yan [13]

Answer:

"The probability that the mean battery life would be greater than 948.8 minutes" is 0.1446.

Step-by-step explanation:

In this case, the quality control expert takes a sample of batteries. From these batteries, we want to find "the probability that the mean battery life would be greater than 948.8 minutes".

Different concepts needed to take into account to solve this question

Sampling Distribution of the Means

For doing this, we need to use the sampling distribution of the means, which results from taking the mean for each possible sample coming from a random variable $\\ x$ . Roughly speaking, each sample will have a different mean, $\\ \overline{x}$ , and the probability distribution for any of these means is called the sampling distribution of the means.

The sampling distribution of the means has a mean that equals the population's mean for the random variable $\\ x$ , i.e., $\\ \mu$ , and its standard deviation is $\\ \frac{\sigma}{\sqrt{n}}$ . We can express this mathematically as:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

Standardized Values for $\\ \overline{x}$

We can standardized the values for $\\ \overline{x}$ using z-scores:

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

This random variable $\\ Z$ follows a standard normal distribution, that is, $\\ Z \sim N(0,1)$ , and it is easier to find probabilities since the values for them are tabulated in the standard normal table (available in any Statistics book or on the Internet.)

What type of distribution follows the sampling distribution of the means?

A general rule of thumb is that this distribution (the sampling distribution of the means) follows a normal distribution if the sample size, $\\ n$ , is bigger than or equal to 30 observations, or $\\ n \geq 30$ . In this case, $\\ n = 109$ batteries. This is a result from the Central Limit Theorem, fundamental in Statistical Inference.

Standard Deviation

We have to remember that the standard deviation is the square root of the variance $\\ \sigma^2$ , or $\\ \sqrt{\sigma^2}$ .

$\\ \sigma^{2} =5929$
$\\ \sigma = \sqrt{5929} = 77$

Therefore, the standard deviation in this case is $\\ \sigma = 77$ minutes.

In sum, we have the following information to answer this question:

$\\ \sigma = 77$ minutes.
$\\ \mu = 941$ minutes.
$\\ n = 109$ batteries (the sample size is large enough to assume that the sampling distribution of the means follows a normal distribution).
$\\ \overline{x} = 948.8$ minutes.

What is the probability that the mean battery life would be greater than 948.8 minutes?

Well, having all the previous information, we can use [2] to solve this question (without using units):

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{948.8 - 941}{\frac{77}{\sqrt{109}}}$

$\\ z = \frac{7.8}{\frac{77}{\sqrt{109}}}$

$\\ z = \frac{7.8}{7.37526}$

$\\ z = 1.05758 \approx 1.06$

This result is the standardized value or z-score for $\\ \overline{x}$ , considering $\\ \mu = 941$ and $\\ \sigma = 77$ .

We round z to two decimals digits since standard normal table only uses it as an entry to find probabilities.

With $\\ z = 1.06$ , we can consult the cumulative standard normal table. First, we need to find with $\\ z = 1.0$ in the first column in the table. Then, in its first raw, we need to find +0.06. The intersection for these two values determines the cumulative probability for $\\ P(z$ .