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nexus9112 [7]
2 years ago
10

Sub to hudson moore among us gameplay

Mathematics
1 answer:
natita [175]2 years ago
3 0

Answer:

okKkAyYy

Step-by-step explanation:

You might be interested in
|x2+x-20<br> 22-11x+28<br> e:<br> x2-x-30<br> 3x-18
elena-s [515]

Answer:

x2 - 10x - 1994

Step-by-step explanation:

Step  1  :

Trying to factor by splitting the middle term

1.1     Factoring  x2-10x-1994

The first term is,  x2  its coefficient is  1 .

The middle term is,  -10x  its coefficient is  -10 .

The last term, "the constant", is  -1994

Step-1 : Multiply the coefficient of the first term by the constant   1 • -1994 = -1994

Step-2 : Find two factors of  -1994  whose sum equals the coefficient of the middle term, which is   -10 .

     -1994    +    1    =    -1993

     -997    +    2    =    -995

     -2    +    997    =    995

     -1    +    1994    =    1993

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Final result :

 x2 - 10x - 1994

Processing ends successfully

plz mark me as brainliest :)

6 0
2 years ago
HELP PLEASE WILL MARK BRAINLIEST TO BEST ANSWER
nlexa [21]

Answer:

c

Step-by-step explanation:

(3x)ex2

3 0
2 years ago
Read 2 more answers
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0
3 years ago
Theresa pays $9.56 for 4 pounds of tomatoes.What is the cost for 1 pound of tomatoe
Andrew [12]
9.56/4 = 2.39

$2.39 per lb.
6 0
3 years ago
Read 2 more answers
The graph below represents the linear equation y=1/2x-3 a second linear equation is represented by the data in the table. What i
stiv31 [10]

ik its late but the correct answer is C:(2,-2)

Step-by-step explanation:

I got the answer correct on Edge 2020.

Hope this helps!!

8 0
3 years ago
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