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aivan3 [116]
3 years ago
6

Tyler used 18 kilograms of potting soil to fill flower boxes. He had already used 23 kilograms of potting soil to fill the flowe

r pots. How many kilograms of potting soil did Tyler use in all?
Mathematics
2 answers:
Nikolay [14]3 years ago
7 0
Tyler used 41 kilograms of potting soil

aksik [14]3 years ago
5 0
41 kilograms Tyler used
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The correct way to set up the slope formula for the line that passes through points (5 , 0) and (6 , -6) is  \frac{-6-0}{6-5} ⇒ C

Step-by-step explanation:

The formula of the slope of a line passes through points (x_{1},y_{1}) and (x_{2},y_{2})

is m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

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∴ x_{1} = 5 and  x_{2} = 6

∴  y_{1} = 0 and  y_{2} = -6

Substitute these values in the formula of the slope

∵ m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

∴ m=\frac{-6-0}{6-5}

Let us look to the answer and find the same formula

The answer is:

The correct way to set up the slope formula for the line that passes through points (5 , 0) and (6 , -6) is  \frac{-6-0}{6-5}

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A school is designed so that 5 hallways meet at the library in the shape of a regular pentagon. Find the measure of one of the e
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A random sample is selected from a normally distributed population. The following sample statistics are obtained: n = 20, = 30,
Flura [38]

Answer:

b) The margin of error is approximately 3.24

e) The critical value is 1.7921.

Step-by-step explanation:

<u>Step(i</u>):-

Given sample size 'n' =20

Given sample standard deviation 's' = 10

<u><em> Margin of error </em></u>

<u><em>The margin of error is determined by</em></u>

<u><em /></u>M.E = \frac{t_{\alpha } S.D }{\sqrt{n} }<u><em /></u>

<em>The level of significance ∝ =0.95</em>

<em>The degrees of freedom = n-1 = 20-1=19</em>

t₀.₉₅ = 1.729

M.E = \frac{1.729 X10 }{\sqrt{20} }

Margin of error = 3.866

Step(ii):-

<u><em> Margin of error </em></u>

<u><em>The margin of error is determined by</em></u>

<u><em /></u>M.E = \frac{t_{\alpha } S.D }{\sqrt{n} }<u><em /></u>

<u><em>Given another sample size n =30</em></u>

<em>The level of significance ∝ =0.95</em>

<em>The degrees of freedom = n-1 = 30-1=29</em>

t₀.₉₅ = 1.70

M.E = \frac{1.7021 X10 }{\sqrt{30} } =3.24

<u><em>Margin of error = 3.24</em></u>

5 0
3 years ago
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