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Vlada [557]
3 years ago
13

A realtor is studying the graph above, which shows the expected value of properties in her area over the next 24 years. If t rep

resents the number of years after 2014, in what year should the increase in property values start to slow down?

Mathematics
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

2030

Step-by-step explanation:

the graph starts to decline 16 years after 2014 so 2014 + 16 = 2030

You might be interested in
. upper left chamber is enlarged, the risk of heart problems is increased. The paper "Left Atrial Size Increases with Body Mass
Sonbull [250]

Answer:

Part 1

(a) 0.28434

(b) 0.43441

(c) 29.9 mm

Part 2

(a) 0.97722

Step-by-step explanation:

There are two questions here. We'll break them into two.

Part 1.

This is a normal distribution problem healthy children having the size of their left atrial diameters normally distributed with

Mean = μ = 26.4 mm

Standard deviation = σ = 4.2 mm

a) proportion of healthy children have left atrial diameters less than 24 mm

P(x < 24)

We first normalize/standardize 24 mm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (24 - 26.4)/4.2 = -0.57

The required probability

P(x < 24) = P(z < -0.57)

We'll use data from the normal probability table for these probabilities

P(x < 24) = P(z < -0.57) = 0.28434

b) proportion of healthy children have left atrial diameters between 25 and 30 mm

P(25 < x < 30)

We first normalize/standardize 25 mm and 30 mm

For 25 mm

z = (x - μ)/σ = (25 - 26.4)/4.2 = -0.33

For 30 mm

z = (x - μ)/σ = (30 - 26.4)/4.2 = 0.86

The required probability

P(25 < x < 30) = P(-0.33 < z < 0.86)

We'll use data from the normal probability table for these probabilities

P(25 < x < 30) = P(-0.33 < z < 0.86)

= P(z < 0.86) - P(z < -0.33)

= 0.80511 - 0.37070 = 0.43441

c) For healthy children, what is the value for which only about 20% have a larger left atrial diameter.

Let the value be x' and its z-score be z'

P(x > x') = P(z > z') = 20% = 0.20

P(z > z') = 1 - P(z ≤ z') = 0.20

P(z ≤ z') = 0.80

Using normal distribution tables

z' = 0.842

z' = (x' - μ)/σ

0.842 = (x' - 26.4)/4.2

x' = 29.9364 = 29.9 mm

Part 2

Population mean = μ = 65 mm

Population Standard deviation = σ = 5 mm

The central limit theory explains that the sampling distribution extracted from this distribution will approximate a normal distribution with

Sample mean = Population mean

¯x = μₓ = μ = 65 mm

Standard deviation of the distribution of sample means = σₓ = (σ/√n)

where n = Sample size = 100

σₓ = (5/√100) = 0.5 mm

So, probability that the sample mean distance ¯x for these 100 will be between 64 and 67 mm = P(64 < x < 67)

We first normalize/standardize 64 mm and 67 mm

For 64 mm

z = (x - μ)/σ = (64 - 65)/0.5 = -2.00

For 67 mm

z = (x - μ)/σ = (67 - 65)/0.5 = 4.00

The required probability

P(64 < x < 67) = P(-2.00 < z < 4.00)

We'll use data from the normal probability table for these probabilities

P(64 < x < 67) = P(-2.00 < z < 4.00)

= P(z < 4.00) - P(z < -2.00)

= 0.99997 - 0.02275 = 0.97722

Hope this Helps!!!

7 0
3 years ago
The diagram shows an empty cylindrical container. Ana puts a solid cube of side length 8 cm into the container. She then pours 1
marshall27 [118]

Answer:

<h3><u>Cylinder</u></h3>

\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}

Given:

  • r = 6 cm
  • h = 18 cm

\begin{aligned}\implies \sf V &= \sf \pi (6)^2(18)\\& = \sf 648 \pi \: cm^3\end{aligned}

<h3><u>Cube</u></h3>

\textsf{Volume of a cube}=\sf x^3\quad\textsf{(where x is the side length)}

Given:

  • x = 8 cm

\begin{aligned}\implies \sf V &= 8^3\\& = \sf 512 \: cm^3\end{aligned}

<h3><u>Volume available to be filled with water</u></h3>

Volume of cylinder - volume of cube

= 684π - 512

= 1532.75204 cm³

    1 litre = 1000 cm³

⇒ 1.5 litres = 1000 × 1.5 = 1500 cm³

As 1500 < 1532.75204, the volume of water poured into the container is smaller than the empty space available in the cylinder.  Therefore, the water will <u>not</u> come over the top of the container.

4 0
2 years ago
Read the quotation from "Song of Myself."
maw [93]

Answer:

the answer is A

4 0
3 years ago
The amount of polonium-210 remaining, p(t), after t days in a sample can be modeled by the exponential function p(t) = 100e−0.00
Archy [21]

Answer:

% Po lost = 100[1 - e^(-0.005t)]  %; 73.0 g

Step-by-step explanation:

p(t) = 100e^(-0.005t)

Initial amount:        p(0) = 100

Amount remaining: p(t) = 100e^(-0.005t)

Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]

% of Po lost  = amount lost/initial amount × 100 %

= [1 - e^(-0.005t)]  × 100 % = 100[1 - e^(-0.005t)]  %

p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g

The mass of polonium remaining after 63 days is 73 g.

4 0
3 years ago
(7+4)6 =<br> Step by step is 42+24=66
ddd [48]

Answer:

66

Step-by-step explanation:

Add what is in the parenthesis, in this case 7+4:

(7+4)

=11

Multiply 11 by 6:

(11)6

=66

hope this helps :)

5 0
3 years ago
Read 2 more answers
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