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2 years ago

756 x 300??????? ?????????

Mathematics

2 answers:

34kurt2 years ago

7 0

Answer:

226,800

Step-by-step explanation:

Calculator

Elan Coil [88]2 years ago

5 0

Answer:226,800

Step-by-step explanation:

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Plsefgrffghttgrewe helppppppppppppppppppppppppppppp

Alborosie

It should be 4 since it is rise over run so it should be 4/1 and that equals positive 4

7 0

2 years ago

Read 2 more answers

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

A car of 1000 kg starts to move from rest and travels a distance of 150 m in 10 second. How much force is needed?

allsm [11]

Initial velocity=u=0m/s
Distance=s=150m
Time=t=10s
Acceleration=a

Using second equation of kinematics

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf\longmapsto 150=0(10)+\dfrac{1}{2}a(10)^2$

$\\ \sf\longmapsto 150=\dfrac{1}{2}\times 100a$

$\\ \sf\longmapsto 50a=150$

$\\ \sf\longmapsto a=\dfrac{150}{50}$

$\\ \sf\longmapsto a=3m/s^2$

Mass=m=1000kg
Force=F

Using newtons second law

$\\ \sf\longmapsto F=ma$

$\\ \sf\longmapsto F=1000(3)$

$\\ \sf\longmapsto F=3000N$

3 0

2 years ago

For the straight line defined by the points (3, 59) and (5, 87), determine the slope and y-intercept. do not round the answers.

Margarita [4]

(3,59)(5,87)
slope = (87 - 59) / (5 - 3) = 28/2 = 14 <== ur slope

y = mx + b
slope(m) = 14
use either of ur points...(3,59)...x = 3 and y = 59
now we sub and find b, the y int
59 = 14(3) + b
59 = 42 + b
59 - 42 = b
17 = b <=== ur y int (0,17)

4 0

2 years ago

For homework Brooke has 15 math problem son social study problems and 5 science problems she has 29 problems in all use mental m

Harlamova29_29 [7]

Answer:

Brooke has 9 problems for social studies.

Step-by-step explanation:

I used addition. I did 15 + 5 = 20. and after that I just knew that you have to add 9 to 20 to get 29, so I figured out that she has 9 social studies problems.

8 0

3 years ago