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Ratling [72]
3 years ago
6

Help me please I don’t get it

Mathematics
1 answer:
eduard3 years ago
4 0

Answer: x - 1

———

x + 2

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The equation 9t = 27.99 models a constant rate situation. HURRY PLZ
MatroZZZ [7]
9t = 27.99
5t = ?
Cross-multiply : 9t? = 139.95t
?=15.55
8 0
3 years ago
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Why might algebra tiles not be a good tool to use to factor x2 + 18x + 80?
kvasek [131]
<span>
factor a trinomial to (10+x) (8+x)
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7 0
3 years ago
A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o
balu736 [363]

Answer:

a

P(X \ge 1) = 0.509

b

P(X  \ge 1) = 0.6807

Step-by-step explanation:

From the question we are told that

   The number of students in the class is  N  =  20  (This is the population )

   The number of student that will cheat is  k =  3

   The number of students that he is focused on is  n  =  4

Generally the probability distribution that defines this question is the  Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally  probability mass function is mathematically represented as

      P(X = x) =  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}

Here C stands for combination , hence we will be making use of the combination functionality in our calculators  

Generally the that  he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

      P(X \ge 1) =  1 - P(X \le 0)

Here  

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{20 - 3} C_{4- 0}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{17} C_{4}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ 1 *  2380}{ 4845}

    P(X \le 0) =  0.491

Hence

    P(X \ge 1) =  1 - 0.491

     P(X \ge 1) = 0.509

Generally the that  he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

    P(X \ge 1) =  1 - P(X \le 0)

   P(X  \ge 1) =1- [  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]

Here n =  6

So

    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{1  *  12376}{38760}]

    P(X  \ge 1) =1- 0.3193

    P(X  \ge 1) = 0.6807

   

5 0
3 years ago
What is the answer to this problem 8-1*0+2/2
tester [92]
8-1*0+2/2
=8-0+1
=9

You use order of operations to solve.
Hope it helped! 
3 0
3 years ago
Read 2 more answers
In a random sample of 9 residents of the state of Florida, the mean waste recycled per person per day was 2.4 pounds with a stan
alina1380 [7]
<h2>Answer with explanation:</h2>

When the sample size is small (< 30) and the population standard deviation is unknown , then we use t-test.

The confidence interval for population mean will be :

\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}   (1)

, where \overline{x} = sample mean

t* = Critical value (based on degree of freedom and significance level).

s= sample standard deviation

n= sample size.

As per given we have

n= 9

Degree of freedom = n-1 = 8

\overline{x}=2.4

s= 0.75

Significance level =\alpha=1-0.80=0.20

Using students' t distribution table ,

Critical value : t^*=t_{\alpha/2,df}=t_{0.10,8}=1.3304

We assume that the population is approximately normal.

Then, a 80% confidence interval for the mean waste recycled per person per day for the population of Florida will be :

2.4\pm (1.3304)\dfrac{0.75}{\sqrt{8}}   (Substitute the values in (1))

2.4\pm (1.3304)\dfrac{0.75}{2.82842712475}

2.4\pm (1.3304)(0.265165042945)

2.4\pm 0.352775573134\approx2.4\pm0.353=(2.4-0.353,\ 2.4+0.353)=(2.047,\ 2.753)

Hence, the 80% confidence interval for the mean waste recycled per person per day for the population of Florida. = (2.047, 2.753)

4 0
3 years ago
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