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4 years ago

25 POINTS!!!!

Mathematics

1 answer:

Kamila [148]4 years ago

8 0

Answer: $509

Step-by-step explanation:

Given : Gabe is getting married in three years and wants to save $20,000 for his wedding. He opens an account earning 6% interest, compounded monthly.

According to the given description, we have

A= $20,000 , r= 6%=0.06 , t= 3 years , n= 12 [1 yearn = 12 months]

Formula to find Periodic monthly deposit :-

$P=\dfrac{A\left(\dfrac{r}{n}\right)}{\left(1+\dfrac{r}{n}\right)^{\left(n\cdot t\right)}-1}$

Substitute given values in the above formula , we get

$P=\dfrac{20000\left(\dfrac{0.06}{12}\right)}{\left(1+\dfrac{0.06}{12}\right)^{\left(3\cdot12\right)}-1}\\\\=\dfrac{100}{1.19668052482-1}\\\\=\dfrac{100}{0.19668052482}=508.438749031\approx509$

Hence, the periodic monthly deposit (payment) needed to achieve a balance of $20,000 after three years= $509.

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The boys mixed 10 gallons of the 20% pure lemon juice and 10 gallons of the 70% pure lemon juice mix, because they wanted 20 gal

mars1129 [50]

The boys should use 140% of pure lemon juice mi to get a final mixture of 20 gallons at 40% lemon juice:

if 20:70 the 40:70 - it is more like ration and proportion.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

6 0

3 years ago

Can someone explain this to me please

Temka [501]

Sin of an angle is opposite leg divided by hypotenuse.

sin B= 12 / 13

tan of an angle is opposite leg divided by adjacent leg

tan B= 12 / 5

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cos B= 5 / 13

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 $\\ P(z$ .

It is important to recall that $\\ P(z$ because $\\ z = 1.06$ is the standardized value for $\\ \overline{x} = 948.8$ minutes.

Then, $\\ P(z$

However, the question is about $\\ P(\overline{x} > 948.8) = P(z>1.06)$

And

$\\ P(\overline{x} > 948.8) + P(\overline{x} < 948.8) = 1$

$\\ P(z>1.06) + P(z$

Then

$\\ P(z>1.06) = 1 - P(z$

$\\ P(z>1.06) = 1 - 0.8554$

$\\ P(z>1.06) = 0.1446$

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

6 0

4 years ago

Zahara asked the students of her class their gymnastic scores and recorded the scores in the table shown below:

Tresset [83]

Answer:

B. 3.5

Step-by-step explanation:

I believe this is correct :')

3 0

2 years ago

There are 32 forwards and 80 guards in Leo's basketball league.

Dmitry_Shevchenko [17]

The greatest common factor is the product of shared primes of the numbers prime factorizations.

32=2*2*2*2*2, 80=2*2*2*2*5 So the GCF is 2*2*2*2=16

So there will be 16 teams with 2 forwards and 5 guards each.

4 0

3 years ago

Read 2 more answers