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expeople1 [14]
2 years ago
13

What integer is the closest solution?

Mathematics
2 answers:
Natasha_Volkova [10]2 years ago
6 0

Answer:

So either the answer will be 8.5

Step-by-step explanation:

It will be 8.5 because if you add a zero at the end of 8.5 that will be 8.50 and 8.50 is half of 9 and 9 is closes to the solution sooo.... thats what i pick

Anna007 [38]2 years ago
6 0
Answer is 8.5 and to explain why it’s 8.5 is because if you add a 0 at the end of 8.5 you will get 8.50 and 8.50 is half of 9 and 9 so that’s the answer
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In the right triangle shown m
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I needa see it tho for I could help
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3 years ago
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Thank you if you do help!:,)
egoroff_w [7]

Answer:

B

Step-by-step explanation:

PEMDAS

8 + 2^3 x 5

8 + 8 x 5

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48

3 0
3 years ago
Nick wanted to determine the length of one blade of a windmill. He stood at a point on the ground 440 feet from the windmill's b
mafiozo [28]

Answer:There are two right triangles. You need to find the opposite side (the height) of each of them. The difference between the two is the height of the blade, x.

You know the adjacent, 440, and you are looking for the opposite, so you need to use tan for both.

tan (30) = y / 440

y = 440 * tan (30) = 254.034...

tan (38.8) = z / 440

y = 440 * tan (38.8) = 353.769.--

353.8 - 254.0 = 99.8 = 100 feet.

Step-by-step explanation:

7 0
1 year ago
Cheri walked 12 blocks in 36 minutes. What was Cheri's speed per minute?
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3 is the speed per minute!
3 0
3 years ago
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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
2 years ago
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