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algol13
3 years ago
15

Consider parallelogram ABCD. If AC¯¯¯¯¯¯¯¯=2x+15 and BD¯¯¯¯¯¯¯¯=5x−12, what does the value of x need to be so that the parallelo

gram is a rectangle?
A: 1
B: -9
C: 9
D: 3/7
Mathematics
1 answer:
Marianna [84]3 years ago
4 0

Answer:

9

Step-by-step explanation:

I took the quick check

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The solution to an any quality is giving in the same building notation as The solution to an inequality is giving in the set bui
Lostsunrise [7]

<u>Options</u>

(A)\left(-\infty , \dfrac23\right]\\\\(B)\left(-\infty , \dfrac23\right) \\\\(C)(\frac23\right, \infty ) \\\\(D) [\frac23\right, \infty )

Answer:

(C)(\frac23\right, \infty )

Step-by-step explanation:

Given the solution to an inequality

{x|x>2/3}

The solution set does not include \dfrac23 , therefore, it must be open at the left. Recall that we use a curvy bracket ( to denote openness at the left.

Since x is greater than  \dfrac23 , the solution set contains all values of larger than  \dfrac23 up till infinity. Since infinity is an arbitrarily large value, we also use an open bracket at the right.

Therefore, another way to represent the solution {x|x>2/3} is:

(\frac23\right, \infty )

The correct option is C.

8 0
3 years ago
. 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
ANSWER ASAP!!
Hoochie [10]

50 FOOT

Step-by-step explanation:

4 0
3 years ago
Which description represents this equation?
ratelena [41]

good zid for me s9ydy9dhchoxyoctd8dgx8yd8yc8yx

4 0
3 years ago
Read 2 more answers
A single card is selected from a deck. find the probability that it is a queen or a black card.
s2008m [1.1K]
52 cards in a deck. 4 of those are queens. 52/2=26 (total # of black cards), 2 red queens. Total + of cards you could possibly draw as the favored car = 28 out of 52...,

This is where i lose the probability. It's greater than 50%. I hope this helps.
8 0
3 years ago
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