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joja [24]
4 years ago
6

Please Please Please answer this sum correctly

Mathematics
2 answers:
alexira [117]4 years ago
7 0
24 pine trees in equal rows, with the number of trees in each row being 10 or more than the number of rows.
Hugo could have planted 12 trees in 2 rows.
zmey [24]4 years ago
7 0

24 pine trees in equal rows, with the number of trees in each row being 10 or more than the number of rows.

Hugo could have planted 12 trees in 2 rows.

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RUDIKE [14]

Answer: gravity affect only the mass

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3 years ago
13-0.75w+ 8x when w = 12and x =1/2
Oksi-84 [34.3K]

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9

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On Monday John bought 7 apples. On Tuesday he bought 3 times as many apples as he did on Monday. How many apples did John buy on
d1i1m1o1n [39]
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What is 2 5/6 +1 5/7
mylen [45]
Hi Cherise1cherhop lets break this equation down with these steps:

1) add the whole numbers first

2) find the LCD of the fractions and that would be 42 since 42 can go into both denominators through multiplication

3) make the denominators (bottom numbers) the same as the LCD (42)

4) simplify it, now the denominators are equal

5) join the denominators together

6) simplify it (3 + 65/42)

7) convert 65/42 to mixed fraction

8) now that we made it a mixed fraction, simplify it

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8 0
4 years ago
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

\mu_s=np=500*0.90=450

The standard deviation will be:

\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

\mu=np=300*0.932=279.6

The standard deviation will be:

\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

P(270\leq x\leq280)=P(269.5

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

P(X=280)=P(279.5

8 0
3 years ago
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