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Ronch [10]
3 years ago
6

stacy's thermostat is set at 74°F.Which range of numbers contains the opposite of stacy's thermostat setting?

Mathematics
1 answer:
viva [34]3 years ago
8 0
I don't have the ranges of numbers to see which range so clearly it's impossible to answer
You might be interested in
g red bell pepper seeds germinates 85% of the time. planted 25 seeds. What is the probability that 20 or more germinate
bixtya [17]

Answer:

P(X\geq 20)= P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)

And replacing using the mass function we got:

P(X=20)=(25C20)(0.85)^{20} (1-0.85)^{25-20}=0.156  

P(X=21)=(25C21)(0.85)^{21} (1-0.85)^{25-21}=0.211  

P(X=22)=(25C22)(0.85)^{22} (1-0.85)^{25-22}=0.217  

P(X=23)=(25C23)(0.85)^{23} (1-0.85)^{25-23}=0.161  

P(X=24)=(25C24)(0.85)^{24} (1-0.85)^{25-24}=0.0759  

P(X=25)=(25C25)(0.85)^{25} (1-0.85)^{25-25}=0.0172  

And adding the values we got:

P(X\geq 20) = 0.8381

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=25, p=0.85)  

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 20)= P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)

And replacing using the mass function we got:

P(X=20)=(25C20)(0.85)^{20} (1-0.85)^{25-20}=0.156  

P(X=21)=(25C21)(0.85)^{21} (1-0.85)^{25-21}=0.211  

P(X=22)=(25C22)(0.85)^{22} (1-0.85)^{25-22}=0.217  

P(X=23)=(25C23)(0.85)^{23} (1-0.85)^{25-23}=0.161  

P(X=24)=(25C24)(0.85)^{24} (1-0.85)^{25-24}=0.0759  

P(X=25)=(25C25)(0.85)^{25} (1-0.85)^{25-25}=0.0172  

And adding the values we got:

P(X\geq 20) = 0.8381

7 0
3 years ago
Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!
IgorC [24]
Hey there, hope I can help!

-4x^2+9y^2+32x+36y-64=0

\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \  9y^2+32x+36y-4x^2=64

\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \  -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4
-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \  -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1

Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1

For me I used
\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1
As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1

Hope this helps!
4 0
3 years ago
-3-2(2x-3) simplified
ddd [48]
First Distribute, then subtract and you’ll get -4x+3
6 0
3 years ago
Solve the system of equations algebraically using substitution and elimination
GrogVix [38]

Answer:

my answer is 2x+ 3y=22 because that the answer

3 0
2 years ago
Two terms of an arithmetic sequence are a10=16 and a35=66 find the sum of the first 75 terms
andrew11 [14]

The sum of the first 75 terms of the arithmetic sequence that has 10th term as 16 and the 35th term as 66 is 5400.  

<h3>How to find the sum of terms using Arithmetic sequence formula</h3>

aₙ = a + (n - 1)d

where

  • a = first term
  • d = common difference
  • n = number of terms

Therefore, let's find a and d

a₁₀ = a + (10 - 1)d

a₃₅ = a + (35 - 1)d

Hence,

16 = a + 9d

66 = a + 34d

25d = 50

d = 50 / 25

d = 2

16 - 9(2) = a

a = 16 - 18

a  = -2

Therefore, let's find the sum of 75 terms of the arithmetic sequence

Sₙ = n / 2 (2a + (n - 1)d)

S₇₅ = 75 / 2 (2(-2) + (75 - 1)2)

S₇₅ = 37.5 (-4 + 148)

S₇₅ = 37.5(144)

S₇₅ = 5400

learn more on arithmetic sequence here: brainly.com/question/1687271

4 0
2 years ago
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