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neonofarm [45]
3 years ago
9

A ​-yard-long field is divided into parts of equal length. Mr. Gump paints a​ 6-yard-long strip in each part. How long is the un

painted strip in each part of the​ field
Mathematics
1 answer:
jok3333 [9.3K]3 years ago
8 0

Answer:

9 yards

Step-by-step explanation:

THE QUESTION ABOVE ISN'T COMPLETE THEN I FIND THE SAME QUESTION WITH JUST CHANGE IN VALUES. CHECK IT BELOW;

A 208-yard-long road is divided into 16 parts of equal length. Mr. Ward paints a 4-yard-long strip in each part. How long is the unpainted strip of each part of the road?

EXPLANATION:

From the question, a 208-yard-long road is been parted into into 16 parts and they all have equal length.

Then we can determine the Length of each part as

(208-yard-long road/16 parts )

= 13

Then each part has a length of 13 yards

From the question, we know that 4-yard-long strip was painted in each part by Mr. Ward. Which means to determine the

unpainted strip of each part, we are going to use the expresion below.

(Length of each strip - length of painted strip)

= (13 yards - 4-yard-long strip )

= 9 yards

Therefore, the length of unpainted strip of each part of the road is 9 yards

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The shorter leg of a right triangle is 6ft shorter than the longer leg. The hypotenuse is 6ft longer than the longer leg. Find t
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Randomly selected students participated in an experiment to test their ability to determine when one minute​ (or sixty​ seconds)
zalisa [80]

Answer: (57.41,\ 62.19)

Step-by-step explanation:

Given : Sample size : n=40

Sample mean : \overline{x}=59.8\text{ seconds}

Standard deviation : \sigma =9.2\text{ seconds}

Significance level : \alpha=1-0.9=0.1

Critical value : z_{\alpha/2}=1.645

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\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=59.8\pm(1.645)\dfrac{9.2}{\sqrt{40}}\\\\\approx59.8\pm2.39\\\\=(59.8-2.39,\ 59.8+2.39)\\\\=(57.41,\ 62.19)

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6 0
3 years ago
The triangle below has a total perimeter of 450 cm:
Korvikt [17]
Equation: 450 = 128 + 150 + n

work:

1st step:
128+150 = 278

2nd step:
450-278 = 172

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3 0
3 years ago
PLEASE HELP!!!
Aliun [14]

Answer:

a. 8.75 M NaOH

b. 0.425 M CuCl₂ or 0.43 M CuCl₂

c. 0.067 M CaCO₃ or 0.07 M CaCO₃

Step-by-step explanation:

Molality is computed using the formula:

M = \dfrac{Moles\:of\:solute}{Liters\:of\:solution}

So first thing you need to do is determine how many moles of solute there are and divide it by the solution in liters.

Converting mass to moles, you need to get the mass of each solute per mole. You can use the periodic table to get the atomic mass (which is the grams per mole of each atom) of each of the elements involved. Then add them up and you will have how many grams per mole of each compound.

1. 35.0g of NaOH in 100ml H₂O

Element         number of atoms            atomic mass           TOTAL

Na                              1                   x             22.99g/mol  =    22.99g/mol

O                                1                   x             16.00g/mol   =    16.00g/mol

H                                1                   x                1.01g/mol   =<u>       1.01g/mol</u>

                                                                                                 40.00g/mol

This means that the molecular mass of NaOH is 40.00 g/mol

Then we use this to convert 35.0g of NaOH to moles:

35.0g \:of\:NaOH \times \dfrac{1\:mole\:of\:NaOH}{40.00g\:of\:NaOH} = \dfrac{35.0\:moles\:of\:NaOH}{40.00}=0.875\:moles\:of\:NaOH

Now that you have the number of moles we divide it by the solution in liters. Before we can do that you have to conver 100ml to L.

100ml\times\dfrac{1L}{1000ml} = 0.1 L

Then we divide it:

\dfrac{0.875\:moles\:of\:NaOH}{0.1L of solution} = 8.75M\: NaOH

2. 20.0g CuCl₂ in 350ml H₂O

Element         number of atoms            atomic mass           TOTAL

Cu                              1                   x             63.55g/mol  =    63.55g/mol

Cl                               2                   x             34.45g/mol   =   <u>70.90g/mol</u>

                                                                                                134.45g/mol

20.g\:of\:CuCl_2\times\dfrac{1\:mole\:of\:CuCl_2}{134.45\:g\:of\:CuCl_2}=0.1488\:moles\:of\:CuCl_2

350ml = 0.350L

\dfrac{0.1488\:moles\:of\:CuCl_2}{0.350L\:of\:solution}=0.425M\:CuCl_2

3. 3.35g CaCO₃ in 500ml

Element         number of atoms            atomic mass           TOTAL

Ca                              1                   x             40.08g/mol  =    40.08g/mol

C                                1                   x              12.01g/mol   =    12.01g/mol

O                                3                  x              16.00g/mol   =<u>   48.00g/mol</u>

                                                                                                100.09g/mol

3.35g\:of\:CaCO_3\times\dfrac{1\:mole\:of\:CaCO_3}{100.09\:g\:of\:CaCO_3}=0.0335\:moles\:of\:CaCO_3

500ml = 0.5L

\dfrac{0.0335\:moles\:of\:CaCO_3}{0.5L}=0.067M\:CaCo_3

4 0
3 years ago
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