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kenny6666 [7]
2 years ago
15

A map has a scale of 1: 500 000.

Mathematics
2 answers:
Shkiper50 [21]2 years ago
8 0

Answer:

350km

Step-by-step explanation:

If a map has a scale of 1: 500000, then 1 cm on the map is 500000 cm

So 7 cm on the map is 35000000 cm = 350 km

Kitty [74]2 years ago
7 0
Answer: 350km

Explanation:

1cm on map = 500000cm on ground
1cm on map = 50km

Then 7cm = 350km
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Divide both sides by x + b to get m by itself. The equation will look like this: m = \frac{y}{x+b}

5 0
3 years ago
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Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean
LuckyWell [14K]

Answer:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

df = n_1 +n_2 -2 = 10+15-2= 23

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

\bar X_1 = 1085 sample mean for group 1

\bar X_2 = 1034 sample mean for group 2

n_1 = 10 sample size for group 1

n_2 = 15 sample size for group 2

s_1 = 52 sample deviation for group 1

s_2 = 61 sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: \mu_1= \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

And the statistic is given by:

t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

And replacing we got:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 10+15-2= 23

And the p value would be:

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0
3 years ago
Suppose a student started with 139.0 mg of trans-cinnamic acid, 463 mg of pyridinium tribromide, and 2.45 mL of glacial acetic a
Sliva [168]

Answer:

0.2889 g  brominated product

64.6 %

Step-by-step explanation:

This is a bromination chemical reaction of an alkene  and we are asked to calculate the theoretical and percent yield. Thus to solve it we have to perform a calculation based on the balanced chemical reaction.

The pyridinium tribromide is used as a generator of molecular bromine in situ, and bromine will add to the double bond in trans-cinnamic acid so we know the reaction occur in a one to one mole fashion.

trans-cinnamic acid + pyridinium  tribromide ⇒ 2,3-dibromo-3-                          

                                                                             phenylpropanoic acid

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mass trans-cinnamic acid  = 139.0 mg x  1g/1000 mg = 0.139 g

# mol trans-cinnamic acid =  0.139 g / 148 g/mol = 9.38 x 10⁻⁴ mol

Since our reaction is 1 mol trans-cinnamic acid  produces 1 mol 2,3-dibromo-3-phenylpropanoic acid, it follows that the theoretical yield is:

1 mol 2,3-dibromo-3-phenylpropanoic acid /  trans-cinnamic acid x 9.38 x 10⁻⁴ mol trans-cinnamic acid  

=  9.38 x 10⁻⁴ mol  2,3-dibromo-3-phenylpropanoic acid

In grams the the theoretical yield is:

molar mass  2,3-dibromo-3-phenylpropanoic acid = 307.97 g/mol

The theoretical mass  2,3-dibromo-3-phenylpropanoic acid:

=   9.38 x 10⁻⁴ mol  2,3-dibromo-3-phenylpropanoic acid x 307.97 g/mol

=   0.2889 g

% yield = mass experimental/mass theoretical

= 0.1866 g /  0.2889 g x 100 = 64.6 %

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