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Gelneren [198K]
3 years ago
15

Find the greatest common factor for each pair of monomials​

Mathematics
1 answer:
hram777 [196]3 years ago
4 0

Answer:

7 :)

Step-by-step explanation:

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A bag contains 6 white marblesw and 4 black marbles.a marble is drawn from the bag and then a second marble is drawn without rep
Katena32 [7]

Answer:

60.5%

Step-by-step explanation:

60% white, 40% black

The first draw you have a 60% chance drawing a white marble, the second time you have a 55% chance since you already drew one in the first round. so i think it is 60.5%% chance drawing white on both rounds since I add 60+55=121. 121/200=60.5%

8 0
3 years ago
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A rectangular picture is 2.5 inches wide and 4 inches long. If the picture is enlarged so that it is 10 inches long, what is the
Andrej [43]
10/2.5=4 So, we should multiply it by 4
4x4=16 width
Pls select brainlest
6 0
3 years ago
In one soccer season, a goalie saved 150 out of
Maru [420]

Answer:

0.81 (rounded to the nearest hundredth) & 81%

Step-by-step explanation:

150/185 = 30/37 ≈ 0.81 (rounded to the nearest hundredth)

0.81 is 81%

7 0
2 years ago
What is the volume of the oblique cube
serious [3.7K]

Answer:

\large\boxed{V=\dfrac{1600\pi}{3}\ cm^3}

Step-by-step explanation:

The formula of a volume of a cone:

V=\dfrac{1}{3}\pi r^2H

r - radius

H - height

We have r = 10cm and H = 16cm. Substitute:

V=\dfrac{1}{3}\pi(10^2)(16)=\dfrac{1}{3}\pi(100)(16)=\dfrac{1600\pi}{3}\ cm^3

6 0
3 years ago
A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21
sammy [17]

Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{340119}{10} = 34011.9

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

S.D = \sqrt{\frac{2711418821}{9}} = 17357.09

Confidence interval:

\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621

34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)

7 0
3 years ago
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