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Kryger [21]
3 years ago
14

Berry Delicious is a popular shop that sells chocolate-covered strawberries. Last year, the shop used 6,540 kilograms of strawbe

rries to make its treats. This year, the shop used 13,080 kilograms. What is the percent of increase in the amount of strawberries used annually?​
Mathematics
1 answer:
Elenna [48]3 years ago
6 0

Answer:

100% increase

Step-by-step explanation:

well first let's find out how much a 100% increase would be. 6,540 + 6,540(when trying to find a 100% change add the number to itself) = 13,080. that gets the answer, so there is no need to continue.

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Scale factor is given by:
(length of larger figure)/(length of smaller figure)=(width of larger figure)/(width of the smaller figure)=3.4
The length of the larger figure will be given by:
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width of the larger figure will be given by:
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Therefore the dimension of the new parallelogram will be 20.4 cm by 15.3 cm
 
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4 years ago
Louise is walking 12 miles to raise funds for a local cause. If she has completed 34 of the distance, how many miles has she wal
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3 years ago
Are the following expressions equivalent 26-(-26) and 26+(-26)
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Step-by-step explanation:

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5 0
3 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

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3 years ago
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