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pashok25 [27]
3 years ago
14

An automotive part manufacturer regularly makes shipments of 400 parts to merks auto repair. On average, 14 of the 400 parts are

found to be defective. How many defective parts should be expected if merk increases his shipment
Mathematics
1 answer:
Radda [10]3 years ago
4 0

Answer:

no matter how many parts are in the shipment, 3.5% of the parts will be defective.

hope this helped, if so hit that thanks button ;D

Step-by-step explanation:


You might be interested in
What is the cost per year of raising a child from birth to 18 years old, and paying for 4 years of private college?
borishaifa [10]

Answer:

345700

Step-by-step explanation:

79600+145700+120400=345700

6 0
4 years ago
A private ship carries two types of cargo: large crates and extra-large crates. Each type of crate weighs a specific amount.
Arisa [49]

Answer:

G. 32

Step-by-step explanation:

32 x 335 = 10720

16,100 - 10720 = 5380/190 = 28.3157894737

275 x 32 = 8800

18,800 - 8800 = 10,000/400 = 25 (This is the closer).

30 x 335 = 10,050

16,100 - 10,050 = 6050/190  = 31.8421052632

30 x 275 = 8250

18,800 - 8250 = 10550/400 = 26.375 (Same reason why it does't work)

25 x 335 = 8375

16,100 - 8375 = 7725/190 = 40.6578947368

25 x 275 = 6875

18,800 - 6875 = 11925/400 = 29.8125

24 x 335 = 8040

16,100 - 8040 = 8060/190 = 42.4210526316

24 x 275 = 6600

18,800 - 6600 = 12200/400 = 30.5

8 0
2 years ago
A 12 foot board is cut so that one piece is 5 times as long as the other. Find the length of each piece.
sveta [45]

Answer:

2 and 10

Step-by-step explanation:

Let x be length of the small piece and y the length of the big one.

● y = 5x

Since the big piece is 5 times longer than the short one.

The total length is 12 ft

● y + x = 12

● 5x + x = 12

● 6x = 12

Divide both sides by 6

● 6x/6 = 12/6

● x = 2

So the length of the two pieces are 2 ft and 10 ft (5×2)

4 0
3 years ago
What are the solutions to this equation?
ziro4ka [17]

 

\displaystyle\\(x-4)^2=49\\\\(x-4)^2-49=0\\\\(x-4)^2-7^2=0\\\\(x-4-7)(x-4+7)=0\\\\(x-11)(x+3)=0\\\\x-11 = 0~~~\text{or}~~~x+3=0\\\\x-11=0~~~\implies~~~\boxed{x_1=11}\\\\x+3=0~~~\implies~~~\boxed{x_2=-3}\\\\\boxed{\bf The~solutions~are:~~x =11~~\text{or}~~x =-3}



8 0
3 years ago
Help please, will mark brainliest ​
mario62 [17]

Answer:

Step-by-step explanation:

LET ORANGES=X

PEARS ARE 3 TIMES =3X

LET CHILDREN=Y

IF WE GIVE 5 ORANGES PER CHILD WE NEED 5Y ORANGES.SO THIS =X

X=5Y........................I

IF WE GIVE 8 PEARS PER CHILD WE NEED 8Y PERARS...THERE ARE STILL 21 PEARS LEFT OUT AFTER THIS THAT IS

8Y+21=3X....................II

SUBSTITUTING FOR X FROM EQN.I..

8Y+21=3*5Y

15Y-8Y=21

7Y=21

Y=3

X==5Y=5*3=15..

SO CHILDREN =3

ORANGES=15

PEARS=3*15=45

THANKYOU AND PLEASE AMRK IT BRAINLIEST

8 0
3 years ago
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