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3 years ago

A rectangular field is 0.3 kilometers long and 0.15 kilometers wide. What is the area of the field in square meters?

Mathematics

1 answer:

Anika [276]3 years ago

6 0

Answer:

4.5 meters

Step-by-step explanation:

The formula for the area of a rectangle is length x width. Using this formula, you can substitute the given information and find the area:

length x width = area

0.3km x 0.15km = area

0.045km = area

Finally, you can now convert square kilometres into square meters by multiplying by 100:

0.045 x 100 = 4.5 meters

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Answer:

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3 years ago

Which of these sugar syrup will be the sweetest ? The sugar syrup with,Required to answer. Single choice.Immersive Reader (1 Poi

aleksley [76]

Answer: c. 150g of sugar dissolved in 300ml water

Step-by-step explanation:

The concentration of sugar syrup will decide which syrup is sweetest. Concentration is defined as the amount of solute dissolved per unit volume of solution.

a) 50g of sugar dissolved in 150ml water

$Concentration=\frac{50g}{150ml}=0.33g/ml$

b) 100g of sugar dissolved in 500ml water

$Concentration=\frac{100g}{500ml}=0.2g/ml$

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$Concentration=\frac{150g}{300ml}=0.5g/ml$

d) 250g of sugar dissolved in 1000ml water

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Thus 150 g of sugar dissolved in 300ml water has highest concentration and is sweetest.

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3 years ago

Determine the intercepts of the line 20 POINTS<br> y-intercept(____,____)<br> x-intercept(____,____)

Ivenika [448]

Answer:

y-intercept(0,0.4)

x-intercept(0.3 ,0)

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3 years ago

Read 2 more answers

When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n whic

Elanso [62]

Answer:

95% Confidence interval for the mean

$142.8 \leq\mu\leq158.4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

$cf=\sqrt{\frac{N-n}{N-1} }$

The standard deviation can be estimated as

$\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963$

The 95% confidence interval has a z value of 1.96, so it becomes:

$M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4$

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3 years ago