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3 years ago

Jean-Patrick found an old unopened bag of lawn fertilizer in his shed. The bag weighs 36 pounds, and the instructions claim

Mathematics

2 answers:

Dovator [93]3 years ago

4 0

Answer: He will need to use 25.2 pounds on his lawn.

Step-by-step explanation: If Jean-Patrick found a bag that contains 36 pounds worth of fertilizer and it can cover 10000 square feet of earth, then he needs to find out how many square feet of earth can be covered by just one pound of fertilizer.

After that, he can multiply that portion by 7000 feet to determine how many pounds in total he would need for his lawn.

To cover one square foot of earth, he would need;

1 ft = 36/10000

1 ft = 0.0036

That means for each square foot of earth, he would need 0.0036 pound of fertilizer. In order to fertilize 7000 square feet of earth, he would need the following

Total = 0.0036 x 7000

Total = 25.2

Jean-Patrick would need 25.2 pounds of fertilizer for his lawn

balandron [24]3 years ago

3 0

Answer:

He needs to use 25.2 pounds to fertilize his lawn.

Step-by-step explanation:

In order to solve this problem we can use a rule of three in such a way that the content of the bag (36 pounds) can cover 10,000 ft² in the same proportion as "x" pounds can cover 7,000 ft², where "x" is the amount of fertilizer he needs to use.

36 pounds -> 10,000 ft²

x pounds -> 7,000 ft²

36/x = 10000/7000

10000*x = 36*7000

x = 36*7000/10000

x = 25.2 pounds

He needs to use 25.2 pounds to fertilize his lawn.

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3 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = the number of defective boards in a random sample of size, n = 25

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= 0.093

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= 0.966

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= 0.034

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$