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Sedaia [141]
3 years ago
10

How can you divide 33/100 by

Mathematics
1 answer:
Darina [25.2K]3 years ago
3 0
You will have to make the fraction a decimal first by dividing the top number by the bottom number then divide if your other number is already a decimal/whole number
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if Penelope's car has an average fuel consumption of 3 miles per gallon and she uses 9.5 gallons how far did she drive ​
KatRina [158]

Answer:

approximately 3.16 repeated

Step-by-step explanation:

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3 years ago
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Eitan randomly selected volcanoes to travel to and study. After traveling, he had seen 13 cinder cone volcanoes, 18 shield volca
Crazy boy [7]

Answer:

6 STRATOVOLCANOES

Step-by-step explanation:

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3 years ago
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Figure A has a perimeter of 24 inches. Figure B has a perimeter of 36 inches and one of its side lengths is 12 inches. Find the
navik [9.2K]

Answer:

12 inches

Step-by-step explanation:

6 0
2 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
3 years ago
You are running a fuel economy study. One of the cars you find is blue. It can travel 38 1/2 miles on 1 1/4 gallons of gasoline.
harkovskaia [24]
Answer:

30.8 miles per gallon

Explanation:

1.25 = 38.5

Divide both sides by 1.25

1 = 30.8
3 0
2 years ago
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