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Advocard [28]
3 years ago
14

If w varies inversely with the square root of t and w = 5 when t = 49,

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

x=k/y

w=k/t

5=k/√49

5=k/7

k=5×7

k=35

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The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .
Misha Larkins [42]

Answer:

0.1507 or 15.07%.

Step-by-step explanation:

We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.

First of all, we will find z-scores for data points using z-score formula.

z=\frac{x-\mu}{\sigma}, where,

z = z-score,

x = Sample score,

\mu = Mean,

\sigma = Standard deviation.

z=\frac{21.97-22}{0.016}

z=\frac{-0.03}{0.016}

z=-0.1875

Let us find z-score of data point 22.03.

z=\frac{22.03-22}{0.016}

z=\frac{0.03}{0.016}

z=0.1875

Using probability formula P(a, we will get:

P(-0.1875

P(-0.1875  

P(-0.1875

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.

6 0
3 years ago
Twice a first number decreased by a second number is 22. The first number increased by four times the second number is - 7.
AnnZ [28]

Answer:

4

Step-by-step explanation:

8 0
3 years ago
Help me pls 20 POINTS
Tema [17]

Answer:

bottom =80m^2

Side =50m^2

triangle = 3 ×8 ×5

6 0
2 years ago
Use each model to predict the life expectancy of residents of a country for which the average annual income is $80,000
Mandarinka [93]

The life expectancies of residents of a country for which the average annual income is $80,000 for the three models are 12309.9352, 172.2436 and 4828.1393

The life expectancies of the models are given as:

y =69.9352 + 0.1530\times (income) --- model 1

y =4.2436 + 0.0021\times (income) --- model 2

y =4.1393 + 0.0603\times (income) --- model 3

Given that the average annual income is $80,000;

We simply substitute 80000 for income in the equations of the three models.

So, we have:

<u>Model 1</u>

y =69.9352 + 0.1530\times (income)

y =69.9352 + 0.1530\times 80000

y =12309.9352

<u>Model 2</u>

y =4.2436 + 0.0021\times (income)

y =4.2436 + 0.0021\times 80000

y =172.2436

<u>Model 3</u>

y =4.1393 + 0.0603\times (income)

y =4.1393 + 0.0603\times 80000

y =4828.1393

Hence, the life expectancies are 12309.9352, 172.2436 and 4828.1393

Read more about linear models at:

brainly.com/question/8609070

4 0
2 years ago
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