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Nadusha1986 [10]
3 years ago
10

Jordy was living in a third story apartment in his complex. Up there he was paying an annual renters insurance premium of $325.0

0. When an opportunity to move to the ground floor came up, he jumped on it, but found out that, since burglaries are more common on ground floor apartments, his renters insurance premium would go up by 18%. What is his new annual renter’s insurance premium. a. $58.50 b. $31.96 c. $268.96 d. $383.50
Mathematics
2 answers:
horrorfan [7]3 years ago
7 0

Answer:

The answer is option D $383.50

Explanation:

Initial annual renters insurance premium = $325

When Jordy moves to ground floor, the premium increases by 18%. So, the amount by which premium increases is = \frac{18}{100}*325= 58.50

So, the new premium amount becomes = 325+58.50 = $383.50

Hence, the answer is $383.50

Savatey [412]3 years ago
4 0
THE CORRECT ANSWER FOR THIS ONE: D. 383.50

HIS RENTERS INSURANCE PREMIUM WOULD GO UP BY 18%. HIS NEW ANNUAL RENTER'S INSURANCE PREMIUM IS NOW 
$383.50
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Which equation(s) below is parallel to y = x-5
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2y= 2x - 10 is the correct answer
8 0
3 years ago
From 19 measurements of a pressure controller, the mean operating pressure is 4.97 kPa with a standard deviation of 0.0461 kPa.
Semenov [28]

Answer:

4.97-2.10\frac{0.0461}{\sqrt{19}}=4.95    

4.97+2.10\frac{0.0461}{\sqrt{19}}=4.99    

So on this case the 95% confidence interval would be given by (4.95;4.99)  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X =4.97 represent the sample mean

\mu population mean (variable of interest)

s=0.0461 represent the sample standard deviation

n=19 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=19-1=18

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that t_{\alpha/2}=2.10

Now we have everything in order to replace into formula (1):

4.97-2.10\frac{0.0461}{\sqrt{19}}=4.95    

4.97+2.10\frac{0.0461}{\sqrt{19}}=4.99    

So on this case the 95% confidence interval would be given by (4.95;4.99)    

5 0
3 years ago
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valina [46]
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6 0
3 years ago
A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playi
Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

<em>Let X = number of names recalled</em>

SO, X ~ N(\mu = 19.41,\sigma^{2} = 3.17^{2})

The z-score probability distribution is given by ;

                  Z = \frac{X-\mu}{\sigma} } } ~ N(0,1)

where, \mu = mean number of names recalled = 19.41

            \sigma = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

     P(X > 15) = P( \frac{X-\mu}{{\sigma} } } > \frac{15-19.41}{3.17}  } ) = P(Z > -1.39)

                                                      = P(Z < 1.39) = 0.9177  {using z table}

<em>Therefore, proportion of the students recalled more than 15 names is </em><em>91.77%.</em>

4 0
3 years ago
Please help!
Ainat [17]
<span>(x – h)^2 + (y – k)^2 = r<span>^2


this equation is a derivative of the equation of a circle

x^2 + y^2 = r^2
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with h = -3 and k = 1

we can plug in each set of numbers and solve.

we find Z to be on the circle edge!</span></span>
8 0
3 years ago
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