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Elena-2011 [213]
2 years ago
7

Courtney knows that there are about 2.2 pounds in every kilogram. If she has a medicine ball

Mathematics
1 answer:
soldi70 [24.7K]2 years ago
3 0

Answer:

For this case we know that the medicine ball weighs 3.6 Kg and we know that there are about 2.2 pounds in every Kg.

So then we can find the amount Kg with this expression:

3.6 Kg *\frac{2.2 pounds}{1 Kg} = 7.92 pounds

And if we round to the nearest integer we got 8 pounds and the best answer would be:

8 pounds

Step-by-step explanation:

For this case we know that the medicine ball weighs 3.6 Kg and we know that there are about 2.2 pounds in every Kg.

So then we can find the amount Kg with this expression:

3.6 Kg *\frac{2.2 pounds}{1 Kg} = 7.92 pounds

And if we round to the nearest integer we got 8 pounds and the best answer would be:

8 pounds

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Answer:

<em>He used </em>1\frac{7}{8}\ lb<em />

<em />

Step-by-step explanation:

Given

Total = 3\ lb

Fraction = \frac{5}{8} to make sauce

Required

Determine the number of pounds used

To solve this, we simply multiply the fraction of tomato used by the total;

Used = Fraction * Total

Used = \frac{5}{8} * 3\ lb

Used = \frac{5  * 3}{8}\ lb

Used = \frac{15}{8}\ lb

Convert to mixed number

Used = 1\frac{7}{8}\ lb

Hence,

<em>He used </em>1\frac{7}{8}\ lb<em> out of the 3lb he bought</em>

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Answer:

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numerator: 1

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Step-by-step explanation:

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70% of the students applying to a university are accepted. Assume the requirements for a binomial experiment are satisfied for 1
marshall27 [118]

Answer:

a) 0.3826

b) 0.9894

Step-by-step explanation:

We are given the following information:

We treat students accepted at university as a success.

P(students accepted at university) = 70% = 0.70

Then the number of students accepted at university follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

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P(x \geq 8) = P(x = 8) + P(x = 9) + P(X=10)\\\\= \binom{10}{8}(0.7)^8(1-0.7)^2 +  \binom{10}{9}(0.7)^9(1-0.7)^1+ \binom{10}{10}(0.7)^{10}(1-0.7)^0\\\\= 0.2334 + 0.1210 + 0.0282\\= 0.3826

b) P(4 or more will be accepted)

P(x \geq 4) =1 - P(x = 0) - P(x = 1) - P(X=2)-P(x=3)\\\\=1 - ( \binom{10}{0}(0.7)^0(1-0.7)^{10} +  \binom{10}{1}(0.7)^1(1-0.7)^9+ \binom{10}{2}(0.7)^{2}(1-0.7)^8+ \binom{10}{3}(0.7)^{3}(1-0.7)^7 )\\\\= 1 - 0.0106\\= 0.9894

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3 years ago
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