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1 year ago

Factor

2 answers:

6 0

The answer would be...

6(x-1)(2x^4+2x^3+2x^2+3x+3)

I hope this helped but I couldn't see this as one of the multiple choice answers but I worked it out and this I what I got

stellarik [79]1 year ago

5 0

C on that one which is the correct answer

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Circle G and X connect at point Y. The length of G Y is 10 centimeters and the length of Y X is 16 centimeters. Point G is the c

poizon [28]

Answer:

its b

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

Addison wants to cut 21 centimeters of lace, but her ruler is only calibrated in millimeters. How many millimeters of lace shoul

serg [7]

1 cm = 10 mm

(21 cm)(10) = 210 mm

5 0

1 year ago

5r^2-44r+120 = -30r+11r

garik1379 [7]

$5r^2-44r+120 = -30r+11r \\ \\ 5f^2 - 44r + 120 = -19r \\ \\ 5r^2 - 44r + 120 + 19r = 0 \\ \\ 5r^2 - 25r + 120 = 0 \\ \\ r = \frac{25+5 \sqrt{71i} }{10} , \frac{25-5 \sqrt{71i} }{10} \\ \\ r = \frac{5+ \sqrt{71i} }{2} , \frac{5 - \sqrt{71i} }{2} \\ \\ ANSWER: r = \frac{5+ \sqrt{71i} }{2} , \frac{5 - \sqrt{71i} }{2}$

3 0

1 year ago

Data was collected for a sample of organic snacks. The amount of sugar (in mg) in each snack is summarized in the histogram belo

castortr0y [4]

Answer:

Step-by-step explanation:

Hello!

The histogram summarizes the amount of sugar in organic snacks. (mg)

Y-axis shows the number of snacks

X-axis shows the amount of sugar per snack

The first column of the histogram show that 2 snacks contain sugar between 220 and 224 mg of sugar.

The second column shows that about 3 snacks have between 224 and 228 mg of sugar.

The third column shows that 8 snacks have between 228 and 232 mg of sugar.

The fourth column shows that 12 snacks have between 232 and 236 mg of sugar.

The fifth column shows that 11 snacks have between 236 and 240 mg of sugar.

The sixth column shows that 3 snacks have between 240 and 244 mg of sugar.

The seventh column shows that 2 snacks have between 244 and 248 mg of sugar.

The total of observations is 2+3+8+12+11+3+2= 41 snacks.

I hope this helps!

3 0

2 years ago

The weight of people in a small town in missouri is known to be normally distributed with a mean of 154 pounds and a standard de

ad-work [718]

If a random sample of 20 persons weighed 3,460, the sample mean x-bar would be 3460/20 = 173 pounds.
The z-score for 173 pounds is given by:
$z=\frac{173-154}{29}=0.655$
Referring to a standard normal distribution table, and using z = 0.66, we find:
$P(\bar x\ \textless \ 173)=0.7454$
Therefore
$P(\bar x\ \textgreater \ 173)=1-0.7454=0.2546$
The answer is: 0.2546

4 0

2 years ago