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timofeeve [1]
2 years ago
14

A particular plant root grows 1.5 inches per month. How many centimeters is the plant root growing per month?

Mathematics
1 answer:
NISA [10]2 years ago
3 0

Answer:3.81 centimeters

Step-by-step explanation:All you have to do is to convert inches into centimeters

So 1 inch=2.54cm

Therefore 1.5inches=(1.5/1)×2.54cm

=3.81cm

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grandymaker [24]

Answer:

The decision made by the researcher based on this information is to reject the null hypothesis.

Step-by-step explanation:

Two-tailed hypothesis test:

Critical value: z_c

Test statistic: z

If |z| < |z_c|, we do not reject the null hypothesis.

If |z| > |z_c|, we reject the null hypothesis.

In this question:

z_c = 1.96, z = -2.05, |z| = 2.05

Since |z| > |z_c|, we reject the null hypothesis.

The decision made by the researcher based on this information is to reject the null hypothesis.

4 0
3 years ago
Use the figure to determine the value of X and y
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Answer:

100,80

Step-by-step explanation:

The opposite angle theorem states that, : "When two lines cross four angles are created and the opposite angles are equal."

y°=80°

Now, check that , x° and y° are in one straight line. A straight line has angle of 180°. So x° and y° should sum up to 180°

x° + y° = 180°

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x° = 100°

6 0
3 years ago
Read 2 more answers
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Burka [1]
The length of the edge would be 5cm
4 0
3 years ago
Last season, your
Katarina [22]

Answer:

55%

Step-by-step explanation:

The percent change formula is written like this:

|\frac{change - original }{original}|. In this scenario, the "change" or new value is 31, because it is this season, and the original is last season. Now, 31-20 = 11, and 11/20 = 55%.

7 0
2 years ago
The equation of a circle in general form is ​ x2+y2+22x+14y−55=0 ​ . What is the equation of the circle in standard form?
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x^2+y^2+22x+14y-55=0 \\\\x^2+22x+y^2+14y=55\\\\(x^2+22x+121)-121+(y^2+14y+49)-49=55\\\\(x+11)^2+(y+7)^2-121-49=55\\\\&#10;(x+11)^2+(y+7)^2-170=55\\\\(x+11)^2+(y+7)^2=55+170\\\\\boxed{(x+11)^2+(y+7)^2=225}

Answer B.
4 0
2 years ago
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