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Greeley [361]
3 years ago
11

What is the order of 5 , -0.1, -5/3 , 0.7, 2 from least to greatest?

Mathematics
1 answer:
Anna11 [10]3 years ago
6 0
If you would like to order the following numbers from least to greatest, you can do this using the following steps:

<span>5, -0.1, -5/3, 0.7, 2
</span>-5/3 = -1.67
<span>
-5/3 < -0.1 < 0.7 < 2 < 5</span>

The correct result would be <span>-5/3 < -0.1 < 0.7 < 2 < 5</span>.
You might be interested in
Binomial probability distributions depend on the number of trials n of a binomial experiment and the probability of success p on
FinnZ [79.3K]

The question is incomplete! Complete question along with answers and step by step explanation is provided below.

Question:

(a) Binomial probability distributions depend on the number of trials n of a binomial experiment and the probability of success p on each trial. Under what conditions is it appropriate to use a normal approximation to the binomial? (Select all that apply.)

nq > 10

np > 5

p > 0.5

np > 10

p < 0.5

nq > 5

(b) What is the probability of "12" or fewer successes for a binomial experiment with 20 trials. The probability of success on a single trial is 0.50. Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

Answer:

(a) The correct options are np > 5 and nq > 5

(b) P(x ≤ 12) = 0.8133

Step-by-step explanation:

Please refer to the attached images for explanation, I am unable to type in text editor due to some technical error!

8 0
3 years ago
Factorise (2t-7)²-(5t-4)² <br>with working pls​
sertanlavr [38]

Answer:

33+12t−21t^2

Step-by-step explanation:

(2t-7)²-(5t-4)²

Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (2t-7)².

4t^2−28t+49−(5t-4)²

Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (5t-4)².

4t^2−28t+49−(25t^2−40t+16)

To find the opposite of 25t^2

−40t+16, find the opposite of each term.

4t^2−28t+49−25t^2−40t+16

Combine 4t^2  and −25t^2  to get −21t^2.

−21t^2−28t+49+40t−16

Combine −28t and 40t to get 12t.

−21t^2+12t+49−16

Subtract 16 from 49 to get 33.

−21t^2+12t+33

Swap terms to the left side.

33+12t−21t^2

I hope this helped!

5 0
3 years ago
What is negation of any number
mamaluj [8]

Answer:

More informally: The two's complement of an integer is exactly the same thing as its negation. ... It means "to find the negation of a number (i.e., its two's complement) you flip every bit then add 1"

7 0
3 years ago
1000 was invested at 5.5% interest, compounded annually. After sometime the amount had grown to 1550. How long was the money inv
scZoUnD [109]

compound interest equation for annually compounded

A=P(1+r)^t

A=final amount

P=principal

r=rate in decimal

t=time in years


given that

A=1550

P=1000

r=5.5%=0.055

find t


1550=1000(1+0.055)^t

divide both sides by 1000

1.55=1.055^t

take ln of both sides

ln(1.55)=ln(1.055^t)

use ln rule ln(a^b)=b(ln(a))

ln(1.55)=t(ln(1.055))

divide both sides by ln(1.055)

\frac{ln(1.55)}{ln(1.055)}=t

using a calculator, we get that t=8.18544 yrs

so about 8.2yrs

4 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
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