When a math question asks you to write an inequality is it the expression I wrote or is it the answer I get?

Samples of a cast aluminum part are classified on the basis of surface finish (in microinches) and edge finish. The results of 1

liubo4ka [24]

Answer:

a) P (A)= 88/102= 0.8627

(b) P (B)= 89/102= 0.8725

c) P (A`) = 14/102 = 0.1372

(d) P (A∩B) = 84/102 =0.8235

(e) P(AUB)= 93/102 = 0.9117

(f)P (A`UB) = 98/102= 0.96078

Step-by-step explanation:

edge

finish excellent good Total

<u>(A )surface finish excellent 84 4 88</u>

good ( B) ⇵ 5 9 14

Total 89 13 102

a) Upper P left-parenthesis Upper A right-parenthesis equals

P (A)= 88/102= 0.8627

All the elements of set A = 84+4= 88

(b) Upper P left-parenthesis Upper B right-parenthesis equals

P (B)= 89/102= 0.8725

All the elements of set B = 84+5= 89

c) Upper P left-parenthesis Upper A prime right-parenthesis equal

P (A`) = 14/102 = 0.1372

All the elements of Universal set U which are not elements of set A = 102- 88= 14

(d) Upper P left-parenthesis Upper A intersection Upper B right-parenthesis equals

P (A∩B) = 84/102 =0.8235

Only those elements of set A and set B which are common

(e) Upper P left-parenthesis Upper A union Upper B right-parenthesis equals

P(AUB)= 93/102 = 0.9117

Totalling elements of set A and B= 88+5= 93

(f) Upper P left-parenthesis Upper A prime union Upper B right-parenthesis equals

P (A`UB) = 98/102= 0.96078

All the elements of Universal set U which are not elements of set A and the elements of Set B = 5+9+ 84= 98

6 0

3 years ago

If Jaime is riding at 14 miles per hour and Cori is riding at 0.40 km per minute, how far will each of them bike in two hours? 1

user100 [1]

Answer:

just take speed =distance ÷time

dustance is speedxtime

14x1.61=22.54÷60 to change to min =0.37km per minute

0.40x0.37= 0.148 km

perhaps I may get wrong

5 0

2 years ago

A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the righ

Naya [18.7K]

Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is = $2.2\times 7.4=16.28$

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is = $2.2\times -6.2=-13.64$

Thus change in moment is: $16.28-(-13.64)=29.92$

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

$\frac{29.92}{0.25}=119.68$

Hence, the constant force exerted on the ball by the wall is 119.68 N.

6 0

3 years ago

I need help with these 2 questions please help me with scie

Andre45 [30]

He's using the graduated cylinder method. I think the volume might be between 15 and 20. (It's kind of not obvious...)

3 0

2 years ago

SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

pav-90 [236]

Answer:

-3f - 10

Step-by-step explanation:

-1 times 10 = -10

-1f times 3f = -3f

Sorry if this isn't right.

4 0

3 years ago

Read 2 more answers