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Gala2k [10]
3 years ago
9

The 8th term of 3, 3/2, 3/4

Mathematics
1 answer:
Goryan [66]3 years ago
7 0

Answer:

3/12

Step-by-step explanation:

You might be interested in
Which fractions are equivalent to -30/48 ?
Vedmedyk [2.9K]

The answer is D

Because

-35/65 = -0.53


I hope that's help:)

6 0
3 years ago
Find (if possible) the complement and the supplement of each angle. (if not possible, enter impossible.) (a) π 10 complement rad
blagie [28]

In this question, we have to find the complement and supplement of the given angles .

Complementary angles are those angles whose sum is 90 degree and supplementary angles are those angles whose sum is 180 degree.

So to find the complement and supplement angles, we need to subtract the given angles from pi/2 and pi respectively .

a.

\frac{ \pi}{10}
\\
complement  \ angle = \frac{ \pi}{2} - \frac{ \pi}{10}     \\                             = \frac{4 \pi}{10} = \frac{ 2 \pi}{5}
\\
Supplement  \ angle = \pi - \frac{ \pi}{10} = \frac{9 \pi}{10}

b.

\frac{9 \pi}{10}
\\
Complement \ angle = \frac{ \pi}{2} - \frac{9 \pi}{10} = \frac{-4 \pi}{10} = -\frac{2 \pi}{5}
\\
Supplement \ angle = \pi - \frac{9 \pi}{10} = \frac{ \pi}{10}

4 0
3 years ago
3. A computer is worth $4000 when it is new. After each year it is worth half what it was the previous year
MrMuchimi

Answer:

Answer is 500

btw did round sorry

Step-by-step explanation:

First year

4,000 divided by 2,000=2,000

Second year

2,000 divided by 2=1,000

Third year

1,000 divided by 2=500

7 0
3 years ago
Read 2 more answers
A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod
tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1                2              3

495      525            470

500         515           480

505        505            460

<u>500         515             470        </u>

<u>∑2000     2060         1880</u>

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4=  470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using  μ ± σ , μ±2 σ  or μ ± 3 σ.

In this question the limits are determined by using  μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4=  470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other  two means are within the given UCL and LCL.

This can be shown by the diagram.

8 0
2 years ago
Plz help!
Marizza181 [45]

Answer:

800 is the answer

Step-by-step explanation:

90-40= 50---->4000/50=800

4 0
2 years ago
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