This is the question I don't understand

How do I prove that a quadrilateral is a rectangle?

maw [93]

I think you want to prove that a rectangle is a quadrilateral. You can prove this with definitions. The definition of a quadrilateral is that it is a shape with four sides. Since a rectangle has four sides, that proves that a rectangle is a quadrilateral.

7 0

3 years ago

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A traffic consultant wants to estimate the true proportion of cars on a certain street that have more than two occupants. She st

neonofarm [45]

Answer:

The data is a simple random sample from the population of interest.

Step-by-step explanation:

Conditions necessary to consider when constructing a confidence interval include using a large sample size, which should be greater Than or equal to 30 for n and a sample proportion of np ≤ 10. Importantly the sample must be a random sample drawn from the population, meaning that all the population of interest must have equal chances of being a part of the sample data. However, the scenariomabiveb, violates this condition as the sampling design is not completely randomized. Those who do not ply the route on that certain day have no Chaves of being part of the sample.

7 0

3 years ago

The value of the x-intercept for the graph of 4x-5y=40 is

Alik [6]

$\hbox{x-intercept - y=0}: \\
4x-5 \times 0=40 \\
4x=40 \\
x=\frac{40}{4} \\
\boxed{x=10}$

8 0

3 years ago

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

RoseWind [281]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

3 years ago

I need help show work<br> ill mark brainliest

Archy [21]

Answer:

The population will be 896 turtles 6 years later ⇒ B

Step-by-step explanation:

The exponential increasing formula is y = a $(1+r)^{x}$ , where

y is the new value

a is the initial value

r is the rate of increase in decimal

x is the time in years

∵ There are 300 turtles

∴ a = 300

∵ The population of the turtles exponentially increases 20% each year

∴ r = 20%

→ Divide it by 100 to change it to decimal

∵ 20% = 20 ÷ 100 = 0.2

∴ r = 0.2

∵ The time is 6 years

∴ x = 6

→ Substitute these values in the exponential formula above

∵ y = 300 $(1+0.2)^{6}$

∴ y = 300 $(1.2)^{6}$

∴ y = 895.7952

→ Round it to the nearest whole number

∴ y = 896

∴ The population will be 896 turtles 6 years later

7 0

3 years ago

This is the question I don't understand​

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