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3 years ago

Solve for x. Each figure is a parallelogram

Mathematics

1 answer:

melomori [17]3 years ago

3 0

Answer:

Step-by-step explanation:

THe sides are the same length

11=x+5

x=6

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Assume that adults have IQ scores that are normally distributed with a mean of 96.296.2 and a standard deviation 20.520.5. Find

Sergeeva-Olga [200]

Answer:

The first quartile of the IQ scores of adults is 82.26.

Step-by-step explanation:

The first quartile is the value that is greater than 25% of the observations and less than 75% of the observation.

The formula to compute the first quartile (Q₁) of a Normal distribution is:

$Q_{1}=\mu-0.68\sigma$

The mean and standard deviation of the IQ scores of adults are:

$\mu=96.2\\\sigma=20.5$

Compute the first quartile value as follows:

$Q_{1}=\mu-0.68\sigma=96.2-(0.68\times20.5)=82.26$

Thus, the first quartile of the IQ scores of adults is 82.26.

5 0

3 years ago

The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation St

saveliy_v [14]

Answer:

a)0.067

b)0.111

c)0.612

d)$687.28

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $385

Standard Deviation, σ = $110

We are given that the distribution of domestic airfares is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(domestic airfare is $550 or more)

P(x > 550)

$P( x > 550) = P( z > \displaystyle\frac{550 - 385}{110}) = P(z > 1.5)$

$= 1 - P(z \leq 1.5)$

Calculation the value from standard normal z table, we have,

$P(x > 550) = 1 - 0.933 = 0.067= 6.7\%$

b) P(domestic airfare is $250 or less)

$P(x \leq 250) = P(z \leq \displaystyle\frac{250-385}{110}) = P(z \leq -1.22)$

Calculating the value from the standard normal table we have,

$P( x \leq 250) = 0.111 = 11.1\%$

c))P(domestic airfare is between $300 and $500)

$P(300 \leq x \leq 500) = P(\displaystyle\frac{300 - 385}{110} \leq z \leq \displaystyle\frac{500-385}{110}) = P(-0.77 \leq z \leq 1.04)\\\\= P(z \leq 1.04) - P(z < -0.77)\\= 0.851 - 0.239 = 0.612 = 61.2\%$

$P(300 \leq x \leq 500) = 61.2\%$

d) P(X=x) = 0.03

We have to find the value of x such that the probability is 0.03.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 385}{110})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 385}{110})=0.03$

$=P( z \leq \displaystyle\frac{x - 385}{110})=0.997$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 385}{110} = 2.748\\x = 687.28$

Hence, the domestic fares must be $687.28 or greater for them to lie in the highest 3%.

7 0

3 years ago

The large rectangle below represents one whole. A large rectangle with 25 equal sections, 11 of which are shaded What percent is

Umnica [9.8K]

fourty four percent hope this helps !

4 0

3 years ago

A sum of 6(3x+8)+32+12x in different ways

Lemur [1.5K]

First you have multiply 6 by both 3x and 8
6*3x = 18x
6*8= 48
18x + 48 + 32 + 12x
now combine like terms
18x + 12x = 30x
48 + 32 = 80
so 30x + 80 is your answer

8 0

3 years ago

Help me pleas this is my 2 try I don't wanna fail

Vika [28.1K]

I hope this helps you

1=3

2=4

1 and 2

3 and 4 adjacent

7 0

3 years ago