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zmey [24]
2 years ago
14

Each month, Nelson pays 0.08 per text message that he sends or receives, plus a $10 fee. Nelson’s bill for February was $44.56.

How many text messages did Nelson send or receive in February?
Mathematics
2 answers:
mamaluj [8]2 years ago
5 0

Answer: 432

Step-by-step explanation: it’s right, pls trust me you’ll get it right because I did for my google form!

insens350 [35]2 years ago
3 0
First, let's set up this problem.

f(x) = 0.08x + 10
How did I get this?
Well, f(x) says that this is a function. The 0.08 comes from the problem. The "x" variable multiplied by the 0.08 comes from how he gets charges 0.08 PER TEXT MESSAGE. The "per" is your key here. Next, he gets charged a FLAT RATE $10 fee.

We can set this up to look like
44.56 = 0.08x + 10
Since he was billed 44.56.
Next, we solve for "x" for the number of texts!

x = 432 texts.
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Nik needs to estimate how many books will fit in a bin. Each book is 1 ft tall, 0.5 ft wide, and 0.1 ft thick. The bin is 5 feet
yawa3891 [41]

Answer:

600 books

Step-by-step explanation:

The bin's dimensions are

5 by 2 by 3

THe volume of the bin is the multiplication of the 3 dimensions given.

Volume of Bin = 5 * 2 * 3 = 30 cubic feet

Now, volume of each book would be gotten the same way. The dimensions of one book is:

1 by 0.5 by 0.1

Volume of 1 book = 1 * 0.5 * 0.1 = 0.05 cubic feet

The number of books that will fit in the bin would be:

30/0.05 = 600 books

6 0
2 years ago
Line l has a slope of 2. The line through which of the following pair of points is perpendicular to l?
djverab [1.8K]

Answer:

D

Step-by-step explanation:

If two lines are perpendicular, the products of their slopes will be -1. Therefore, we're looking for a line with slope of -1/2. Let's check each answer:

A: 5 - 3 / (-2 - 3) = 2 / -5    

B: 6 - 5 / 6 - 5 = 1

C: 5 - 3 / 4 - 3 = 2

D: 4 - 3 / 3 - 5 = -1/2

The answer is D.

8 0
2 years ago
Read 2 more answers
A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take
ddd [48]

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

7 0
2 years ago
"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt
Westkost [7]

Answer:

a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c) P(\bar X >31.25)=0.006=0.6\%

d) P(\bar X >28.25)=0.9997=99.97\%

e) P(28.25

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =2)

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b. What is the standard error of the mean time?

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >31.25)=1-P(\bar X

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >28.25)=1-P(\bar X

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

P(28.25

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2 years ago
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