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Y_Kistochka [10]
3 years ago
7

Two friends are renting an apartment. They pay the landlord the first month’s rent. The landlord also requires them to pay an ad

ditional half of a month’s rent for a security deposit. The total amount they pay the landlord before moving in is $1725. What is the monthly rent?
A.
x−1/2x=1725; $3450


B.
x = 1725; $1725


C.
x+1/2x=1725; $1150


D.
x+1/2x=1725; $575
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0
C. x+1/2x=1725; $1150
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The average THC content of marijuana sold on the street is 9.3%. Suppose the THC content is normally distributed with standard d
velikii [3]

Answer:

a) X \sim N(9.3,1)  

b) P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z

c) The value of height that separates the bottom 75% of data from the top 25% is 9.9745.  

Step-by-step explanation:

1) Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

X \sim N(9.3,1)  

Where \mu=9.3 and \sigma=1

3) Part b

We are interested on this probability

P(X>9.2)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z

4) Part c

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.6745. On this case P(Z<0.6745)=0.75 and P(z>0.6745)=0.25

If we use condition (b) from previous we have this:

P(X  

P(Z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.6745

And if we solve for a we got

a=9.3 +1*0.6745=9.9745

So the value of height that separates the bottom 75% of data from the top 25% is 9.9745.  

8 0
3 years ago
Which pair of numbers are equivalent? 0.65% and 65100.65% and 65 tenths 0.7 and 7% 0.7 and 7% 0.5 and 150.5 and 1 fifth 0.02 and
N76 [4]

Step-by-step explanation:

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6 0
3 years ago
Deborah has $56 in her bank account currently. She earns $8 per hour at her job at the local ice cream shop. Micah has $32 in hi
Scilla [17]
5 hours I did this on Plato
4 0
2 years ago
1) If the alpha level is changed from α = .05 to α = .01, what happens to boundaries for the critical region?
Alexxandr [17]

Answer:

1) a. Move farther into the tails

2) a. Decreases

Step-by-step explanation:

Hello!

1)

Let's say for example that you are making a confidence interval for the mean, using the Z-distribution:

X[bar] ± Z_{1-\alpha /2} * \frac{Sigma}{\sqrt{n} }

Leaving all other terms constant, this are the Z-values for three different confidence levels:

90% Z_{0.95}= 1.648

95% Z_{0.975}= 1.965

99% Z_{0.995}= 2.586

Semiamplitude of the interval is

d= Z_{1-\alpha /2} * \frac{Sigma}{\sqrt{n} }

Then if you increase the confidence level, the value of Z increases and so does the semiamplitude and amplitude of the interval:

↑d= ↑Z_{1-\alpha /2} * \frac{Sigma}{\sqrt{n} }

They have a direct relationship.

So if you change α: 0.05 to α: 0.01, then the confidence level 1-α increases from 0.95 to 0.99, and the boundaries move farther into the tails.

2)

The significance level of a hypothesis test is the probability of committing a Type I error.

If you decrease the level from 5% to 1%, then logically, the probability decreases.

I hope this helps!

5 0
3 years ago
Find the z-score that has 62.8% of the distribution's area to its right.
horrorfan [7]
<span>The best and easiest way to get it is by using a TI calculator and i hope you can have it, then you can type in the next: invnorm(1 - 0.628) What you are doing is that you are subtracting the probability from 1 because the invnorm function finds the z-score given the probability to the left of the z-score. I think that can help you. </span>

8 0
3 years ago
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