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Nesterboy [21]
3 years ago
7

zena buys 8 pieces of fruit. she buys fewer cantaloupes than apples and more apples than pears . she buys 3 pears. how many cant

aloupes and how many apples did she buy
Mathematics
1 answer:
Pie3 years ago
5 0
8 pieces of fruit and 3 pears
means she bought 4 apples and 1 cantaluope
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A set of equations is given below: Equation C: y = 6x + 9 Equation D: y = 6x + 2 Which of the following best describes the numbe
7nadin3 [17]
So we can get from these two equations that y = 6x + 2 = 6x + 9

6x + 2 = 6x + 9
0 = 7 , but that's incorrect, so this set has no solution.
7 0
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432,2=26<br>275,3=66<br>342,5=41<br>123,4=20<br>then <br>543,2=?​
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Answer:

I think 543,2=24.

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3 years ago
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natita [175]
5 the answer I took the test
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A multiple-choice standard test contains total of 25 questions, each with four answers. Assume that a student just guesses on ea
weqwewe [10]

Answer:

a)  8*88*10⁻⁶ ( 0.00088 %)

b) 0.2137 (21.37%)

Step-by-step explanation:

if the test contains 25 questions and each questions is independent of the others, then the random variable X= answer "x" questions correctly , has a binomial probability distribution. Then

P(X=x)= n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of questions= 25

p= probability of getting a question right = 1/4

then

a) P(x=n) = p^n = (1/4)²⁵ = 8*88*10⁻⁶ ( 0.00088 %)

b) P(x<5)= F(5)

where F(x) is the cumulative binomial probability distribution- Then from tables

P(x<5)= F(5)= 0.2137 (21.37%)

7 0
3 years ago
Evaluate the sum of the following finite geometric series.
rjkz [21]

Answer:

\large\boxed{\dfrac{156}{125}\approx1.2}

Step-by-step explanation:

<h3>Method 1:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}

<h3>Method 2:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}

a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1

\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}

5 0
3 years ago
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