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Margaret [11]
2 years ago
11

I just took my first sat practice test and scored a 1310, I’ve only been studying for a week, but I still feel a little disappoi

nted. Is this a good score? I’m a junior btw
SAT
1 answer:
quester [9]2 years ago
6 0
That is a very good score! Don’t be disappointed. It’s okay to retake the SAT if you are not happy with the score. Most colleges take your highest score anyways. Good job!
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What does the underlined word mean in the following sentence? Mónica es hermosa. ugly mad nice beautiful
natita [175]

Answer:

it's <u><em>beautiful.</em></u>

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3 years ago
Can someone pls help me with latin??
Katen [24]
What’s the question?????
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3 years ago
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A benefit of the Direct Stafford Loan is that:
Ratling [72]

One of the core benefit of the Direct Stafford Loan is that: A. it doesn't accrue interest.

<h3>What is a student loan?</h3>

A student loan is a governmental loan or a loan that's obtained from a private lender, so as to avail a student the ability to pay his or her college fees such as:

  • Tuitions
  • Books
  • Living costs
  • Supplies

In this context, one of the core benefit of the Direct Stafford Loan is that it doesn't accrue interest because it's fixed for the life of the loan.

Read more on student loan here: brainly.com/question/16724065

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6 0
2 years ago
Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a standard deviation of 1500 hours
FromTheMoon [43]

Answer:

0.2776

Explanation:

Mean = 15000, SD = 1500

We need to find Cumulative probability: P(X > 15884)

First we need to convert it into normal distribution.

From the attached file, we can see the shaded area we are looking for.

For conversion as follow; P (X>15884) =P ( X - mean > 15884-15000 ) =P (\frac{X- mean}{SD} > \frac{15884-15000}{1500} )

Next to find Z = \frac{X- mean}{SD} = 0.59

now we have converted to normal distribution and found the value of Z, we need to find the probability P(X > 15884).

P(X > 15884) = P(Z > 0.59).

From the normal distribution table at Z= 0.59, and greater than 0.59, Probability = 0.2776.

4 0
3 years ago
Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enroll
Andrew [12]

Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm t\frac{s}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed <em>95% confidence interval</em>, with 49 - 1 = <em>48 df</em>, is t = 2.0106.

Hence:

\overline{x} - t\frac{s}{\sqrt{n}} = 12.2 - 2.0106\frac{1.6}{\sqrt{49}} = 11.7

\overline{x} + t\frac{s}{\sqrt{n}} = 12.2 + 2.0106\frac{1.6}{\sqrt{49}} = 12.7

The 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).

More can be learned about the t-distribution at brainly.com/question/16162795

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2 years ago
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