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3 years ago

20 points!

Mathematics

2 answers:

meriva3 years ago

4 0

All circles are similar because all circle have same shape.

However Two circles are congruent , if and only if their radii are congruent.

In our question we are asked when are all circles similar given if their radii are congruent.

So we can say this statement is false because no matter the radii are congruent or not , two or more circles are always similar because of their same shape no matter what the measure of their radii is. Only if we are asked when they are congruent, we will consider the radii part.

Answer is false.

anyanavicka [17]3 years ago

4 0

False
All circles are similar because every circle is a dilation of another circle.

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jeka94

Multiply both sides by lh, V=wlh

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2 years ago

Read 2 more answers

16x2 + 10x – 27 = -6x + 5<br> What are the solutions to this equation ?<br> X<br> x=

bonufazy [111]

Answer:

x=-2 x=1

Step-by-step explanation:

16x2 + 10x – 27 = -6x + 5

Add 6x to each side

16x^2+6x + 10x – 27 = -6x+6x + 5

16x^2 +16x -27 = 5

Subtract 5 from each side

16x^2 + 16x – 27-5 = 5 - 5

16x^2 +16x -32 = 0

Factor out 16

16 (x^2 +x-2)=0

Factor

16 (x+2) (x-1) =0

Using the zero product property

(x+2) =0 x-1=0

x=-2 x=1

5 0

3 years ago

A study conducted at a certain high school shows that 72% of its graduates enroll at a college. Find the probability that among

mixas84 [53]

Answer:

$P(X \geq 1) =1-P(X$

And we can use the probability mass function and we got:

$P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615$

And replacing we got:

$P(X \geq 1) = 1-0.00615 = 0.99385$

Step-by-step explanation:

Let X the random variable of interest "number of graduates who enroll in college", on this case we now that:

$X \sim Binom(n=4, p=0.72)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We want to find the following probability:

$P(X \geq 1)$

And we can use the complement rule and we got:

$P(X \geq 1) =1-P(X$

And we can use the probability mass function and we got:

$P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615$

And replacing we got:

$P(X \geq 1) = 1-0.00615 = 0.99385$

5 0

3 years ago

Math please help me thank you :)

Pavlova-9 [17]

I think your answer is ( B )

5 0

2 years ago

The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$

To convert billion to million you multiple by 1000

$P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year$

6 0

2 years ago