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Finger [1]
3 years ago
13

30. TRAVEL Before starting her vacation, Sari's car had 23,500 miles on the

Mathematics
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

(23,500 +14m) miles

Step-by-step explanation:

the mileage on Sari's odometer after her trip =

(23,500 + 14 m) miles

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NEED HELP ASAP I DONT GET THIS
denpristay [2]

Answer:

x = 60

Step-by-step explanation:

In a circle, the sum of the angles that meet at a point is 360 degrees. So now that we have the total, we can subtract all the given information in the picture. We have 84, 75, 68, and x + 73. Now let us set up and equation for this problem.

360 = 84 + 75 + 68 + x + 73

Now combine like terms.

360 = x + 300

Now subtract 300 on both sides.

x = 60

And this is the value of x.

5 0
3 years ago
Read 2 more answers
g 78% of all Millennials drink Starbucks coffee at least once a week. Suppose a random sample of 50 Millennials will be selected
Tatiana [17]

Answer:

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

np=50*0.78=39  \geq 10

n(1-p)=50*(1-0.78)=11 \geq 10

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

p \sim N (p, \sqrt{\frac{p(1-p)}{n}})

With the following parameters:

\mu_ p = 0.78

\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

np=50*0.78=39  \geq 10

n(1-p)=50*(1-0.78)=11 \geq 10

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

p \sim N (p, \sqrt{\frac{p(1-p)}{n}})

With the following parameters:

\mu_ p = 0.78

\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586

6 0
3 years ago
a music store brought a CD set at a cost of $10. When the store sold the CD set,the precent markup was 10%. Find the selling pri
saw5 [17]

Answer:

1

Step-by-step explanation:

10% percent of 10 is 1.   I hope this helps you out!

4 0
3 years ago
Read 2 more answers
Danielle wants to paint her daughter's room pink. Which plan will give her paint that is the least red?
NeTakaya

Step-by-step explanation:

Plan B:

2:5 = 0.4 red to 1 white

1:4 = 0.25 red to 1 white

Students might also scale up using part-to-whole-ratios and come up with 35 parts total and 14 parts red for Plan A and 35 parts total and 7 parts red for Plan B.

8 0
3 years ago
A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G
dezoksy [38]

Answer:

a) The mean of a sampling distribution of \\ \overline{x} is \\ \mu_{\overline{x}} = \mu = 20. The standard deviation is \\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2.

b) The standard normal z-score corresponding to a value of \\ \overline{x} = 16 is \\ Z = -2.

c) The standard normal z-score corresponding to a value of \\ \overline{x} = 23 is \\ Z = 1.5.

d) The probability \\ P(\overline{x}.

e) The probability \\ P(\overline{x}>23) = 1 - P(Z.

f)  \\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z.

Step-by-step explanation:

We are dealing here with the concept of <em>a sampling distribution</em>, that is, the distribution of the sample means \\ \overline{x}.

We know that for this kind of distribution we need, at least, that the sample size must be \\ n \geq 30 observations, to establish that:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the distribution of the sample means follows, approximately, a <em>normal distribution</em> with mean, \mu, and standard deviation (called <em>standard error</em>), \\ \frac{\sigma}{\sqrt{n}}.

The number of observations is n = 64.

We need also to remember that the random variable Z follows a <em>standard normal distribution</em> with \\ \mu = 0 and \\ \sigma = 1.

\\ Z \sim N(0, 1)

The variable Z is

\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of \\ \overline{x} is the population mean \\ \mu = 20 or \\ \mu_{\overline{x}} = \mu = 20.

The standard deviation is the population standard deviation \\ \sigma = 16 divided by the root square of n, that is, the number of observations of the sample. Thus, \\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2.

Part b

We are dealing here with a <em>random sample</em>. The z-score for the sampling distribution of \\ \overline{x} is given by [1]. Then

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

\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}

\\ Z = \frac{-4}{\frac{16}{8}}

\\ Z = \frac{-4}{2}

\\ Z = -2

Then, the <em>standard normal z-score</em> corresponding to a value of \\ \overline{x} = 16 is \\ Z = -2.

Part c

We can follow the same procedure as before. Then

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

\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}

\\ Z = \frac{3}{\frac{16}{8}}

\\ Z = \frac{3}{2}

\\ Z = 1.5

As a result, the <em>standard normal z-score</em> corresponding to a value of \\ \overline{x} = 23 is \\ Z = 1.5.

Part d

Since we know from [1] that the random variable follows a <em>standard normal distribution</em>, we can consult the <em>cumulative standard normal table</em> for the corresponding \\ \overline{x} already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is <em>two standard units</em> <em>below</em> the mean (because of the <em>negative</em> value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is \\ P(Z.

Therefore, the probability \\ P(\overline{x}.

Part e

We can follow a similar way than the previous step.

\\ P(\overline{x} > 23) = P(Z > 1.5)

For \\ P(Z > 1.5) using the <em>cumulative standard normal table</em>, we can find this probability knowing that

\\ P(Z1.5) = 1

\\ P(Z>1.5) = 1 - P(Z

Thus

\\ P(Z>1.5) = 1 - 0.9332

\\ P(Z>1.5) = 0.0668

Therefore, the probability \\ P(\overline{x}>23) = 1 - P(Z.

Part f

This probability is \\ P(\overline{x} > 16) and \\ P(\overline{x} < 23).

For finding this, we need to subtract the cumulative probabilities for \\ P(\overline{x} < 16) and \\ P(\overline{x} < 23)

Using the previous <em>standardized values</em> for them, we have from <em>Part d</em>:

\\ P(\overline{x}

We know from <em>Part e</em> that

\\ P(\overline{x} > 23) = P(Z>1.5) = 1 - P(Z

\\ P(\overline{x} < 23) = P(Z1.5)

\\ P(\overline{x} < 23) = P(Z

\\ P(\overline{x} < 23) = P(Z

Therefore, \\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z.

5 0
3 years ago
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