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nexus9112 [7]
3 years ago
15

please help me with this math problem, the question is on the picture that i attached, ill mark brainliest if the answer is corr

ect. NO links!

Mathematics
1 answer:
dsp733 years ago
7 0

Answer:

cos z = 35 / 37

Step-by-step explanation:

cos z is the ratio of the adjacent side of a triangle (the side adjacent to angle z to the hypotenuse:

cos z = (adj. side) / (hypotenuse)

Here we have cos z = 35 / 37

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A financial advisor is analyzing a family's estate plan. The amount of money that the family has invested in different real esta
Pachacha [2.7K]

Answer:

The amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of money that the family has invested in different real estate properties.

The random variable <em>X</em> follows a Normal distribution with parameters <em>μ</em> = $225,000 and <em>σ</em> = $50,000.

It is provided that the family has invested in <em>n</em> = 10 different real estate properties.

Then the mean and standard deviation of amount of money that the family has invested in these 10 different real estate properties is:

\mu_{\bar x}=\mu=\$225,000\\\\\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{50000}{\sqrt{10}}=15811.39

Now the lowest 80% of the amount invested can be represented as follows:

P(\bar X

The value of <em>z</em> is 0.84.

*Use a <em>z</em>-table.

Compute the value of the mean amount invested as follows:

\bar x=\mu_{\bar x}+z\cdot \sigma_{\bar x}

   =225000+(0.84\times 15811.39)\\\\=225000+13281.5676\\\\=238281.5676\\\\\approx 238281.57

Thus, the amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

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3 years ago
N − 7.32 / 2 = 1.04<br><br> What is "n"?
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n= 4.7

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n+−3.66=1.04

n−3.66=1.04

Step 2: Add 3.66 to both sides.

n−3.66+3.66=1.04+3.66

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n=4.7

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One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independ
Illusion [34]

Question has missing details

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is 0.05. What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly?

Answer:

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

Step-by-step explanation:

Given

Let P = Probability that a point signals an error incorrectly = 0.05

Let Q = Probability that a point signals an error correctly

P + Q = 1 ---- Make Q the subject of formula

Q = 1 - P where P = 0.05

So, Q = 1 - P becomes

Q = 1 - 0.05

Q= 0.95

Solving for the probability that at least one of 10 successive points indicates a problem when in fact the process is operating.

If two events (P and Q) are independent

Then

P(P n Q) = P(P) * P(Q)

From De Morgan law;

P(P u Q) = 1 - P(P' n Q')

Where P(P u Q) represent the probability that at least one of 10 successive points

P(P' n Q') is calculated as follows;

P(P' n Q') = 0.95^10

P(P' n Q') = 0.59873693923837890625

So,

P(P u Q) = 1 - P(P' n Q') becomes

P(P u Q) = 1 - 0.59873693923837890625

P(P u Q) = 0.40126306076162109375

P(P u Q) = 0.4013 ----- Approximated

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

7 0
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