GIVING BRAINLIEST PLEASE ANSWER

Mathematics

2 answers:

slava [35]3 years ago

8 0

Answer:

you would have 13 peoples in the Eiffel Tower side of the diagram. and then you would have 16 people on the Notre Dame cathedral side. and none in the middle of the diagram. the answer to question b would 29/30 since that is the number of people who exclusively went to et or ndc.

Step-by-step explanation:

if that isnt the right answer is it possible the question was phrased incorrectly and instead was supposed to 7 people went to both et and ndc?

klasskru [66]3 years ago

7 0

Answer:

Step-by-step explanation:

aww im sorry i dont no i wish i knew but you dont need to give me anything im stupid

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6,000 multiplied by power of 10

larisa86 [58]

An easy way of finding the answer to your question is by using a scientific calculator. When I plugged in 6,000^10 it gave me 6.0466176x10^37. Yes the answer is in scientific notation, but if you want the answer shortened out to the nearest hundredth place the answer is 6.05.

7 0

3 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:

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