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Aleksandr [31]
3 years ago
7

PLEASE HURRY AND ANSWER FOR ME

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
5 0
It’s -4 and it’s -5.........
DiKsa [7]3 years ago
5 0
6 is bigger than -30
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Aircraft landing gears coming from a facility are inspected for defects with the aim to improve the failure rates. Historical da
erica [24]

Answer:

78% probability it contains no defects

Step-by-step explanation:

We have that:

10% have defects in shafts only.

8% have defects in bushings only.

4% have the two types of defects.

An assembly is selected arandom, what is the probability it contains no defects

Either it has at least one defect, or it has no defects. The sum of these probabilities is 100%. So

D + ND = 100

Probability that it has a defect

D = 10 + 8 +4 = 22%

No defects:

22 + ND = 100

ND = 78

78% probability it contains no defects

5 0
3 years ago
3, 9, 15, ...<br> Find the 45th term.
Lelechka [254]
First the nth term which is 6n-3
So 6x45-3 is your answer
8 0
3 years ago
Read 2 more answers
On a road trip in Motanna, USA, Elise sees a road sign that says Helena 87. Elise tests the accuracy of her cars odometer and tr
mote1985 [20]

Answer:

No i dont think so but im not sure

Step-by-step explanation:


8 0
3 years ago
What is the volume of a right circular cone that has a height of 14.9 cm and a base with the circumference of 2.9 cm rounded to
xxMikexx [17]

Answer:

The answer to your question is  Volume = 3.30 cm³

Step-by-step explanation:

Data

Volume = ?

height = 14.9 cm

circumference = 2.9 cm

Process

1.- Calculate the radius

   circumference = 2πr

-Substitution

   2.9 = 2πr

-Solve for r

   r = 2.9/2π

   r = 1.45/π cm

2) Calculate the volume of the cone

   Volume = 1/3πr²h

-Substitution

   Volume = 1/3π(1.45/π)²(14.9)

-Simplify

   Volume = (1.04)(0.213)(14.9)

   Volume = 3.30 cm³

5 0
3 years ago
I need help with this problem please
NNADVOKAT [17]

<em>n</em> must be 0, since <em>x </em>ⁿ = 1 for all positive, real <em>x</em> if <em>n</em> = 0. So the answer is A.

7 0
3 years ago
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